2.2: Functions of Multiple Variables
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Our first step is to explain what a function of more than one variable is, starting with functions of two independent variables. This step includes identifying the domain and range of such functions and learning how to graph them. We also examine ways to relate the graphs of functions in three dimensions to graphs of more familiar planar functions.
Functions of Two Variables
The definition of a function of two variables is very similar to the definition for a function of one variable. The main difference is that, instead of mapping values of one variable to values of another variable, we map ordered pairs of variables to another variable.
Definition: Function of Two Variables
A function of two variables \(z=f(x,y)\) maps each ordered pair \((x,y)\) in a subset \(D\) of the real plane \(\rm I\!R^2\) to a unique real number \(z\). The set \(D\) is called the domain of the function. The range of \(f\) is the set of all real numbers \(z\) that has at least one ordered pair \((x,y)∈D\) such that \(f(x,y)=z\) as shown in Figure \(\PageIndex{1}\).
Determining the domain of a function of two variables involves taking into account any domain restrictions that may exist. Let’s take a look.
Example \(\PageIndex{1}\): Domains and Ranges for Functions of Two Variables
Find the domain and range of each of the following functions:
- \(f(x,y)=3x+5y+2\)
- \(g(x,y)=\sqrt{9−x^2−y^2}\)
Solution
a. This is an example of a linear function in two variables. There are no values or combinations of \(x\) and \(y\) that cause \(f(x,y)\) to be undefined, so the domain of \(f\) is \(\rm I\!R^2\). Written in set-builder notation, this could be written as, \(\{ (x, y) | x \in \rm I\!R, y \in \rm I\!R \}\).
To determine the range, first pick a value for z . We need to find a solution to the equation \(f(x,y)=z,\) or \(3x−5y+2=z.\) One such solution can be obtained by first setting \(y=0\), which yields the equation \(3x+2=z\). The solution to this equation is \(x=\dfrac{z−2}{3}\), which gives the ordered pair \(\left(\dfrac{z−2}{3},0\right)\) as a solution to the equation \(f(x,y)=z\) for any value of \(z\). Therefore, the range of the function is all real numbers, or \((-\infty, \infty)\).
b. For the function \(g(x,y)\) to have a real value, the quantity under the square root must be nonnegative:
\[ 9−x^2−y^2≥0. \nonumber\]
This inequality can be written in the form
\[ x^2+y^2≤9. \nonumber\]
Therefore, the domain of \(g(x,y)\) is \(\{(x,y)∈R^2∣x^2+y^2≤9\}\). The graph of this set of points can be described as a disk of radius 3 centered at the origin. The domain includes the boundary circle as shown in the following graph.
To determine the range of \(g(x,y)=\sqrt{9−x^2−y^2}\) we start with a point \((x_0,y_0)\) on the boundary of the domain, which is defined by the relation \(x^2+y^2=9\). It follows that \(x^2_0+y^2_0=9\) and
\[ \begin{align*} g(x_0,y_0) &=\sqrt{9−x^2_0−y^2_0} \\[5pt]&=\sqrt{9−(x^2_0+y^2_0)}\\[5pt]&=\sqrt{9−9}\\[5pt]&=0. \end{align*}\]
If \(x_0^2+y_0^2=0\) (in other words, \(x_0=y_0=0)\), then
\[ \begin{align*} g(x_0,y_0)&=\sqrt{9−x^2_0−y^2_0}\\[5pt]&=\sqrt{9−(x^2_0+y^2_0)}\\[5pt]&=\sqrt{9−0}=3. \end{align*}\]
This is the maximum value of the function. Given any value \(c\) between \(0\) and \(3\), we can find an entire set of points inside the domain of \(g\) such that \(g(x,y)=c:\)
\[\begin{align*} \sqrt{9−x^2−y^2}&=c \\[5pt] 9−x^2−y^2&=c^2 \\[5pt] x^2+y^2&=9−c^2. \end{align*}\]
Since \(9−c^2>0\), this describes a circle of radius \(\sqrt{9−c^2}\) centered at the origin. Any point on this circle satisfies the equation \(g(x,y)=c\). Therefore, the range of this function can be written in interval notation as \([0,3].\)
Exercise \(\PageIndex{1}\)
Find the domain and range of the function \(f(x,y)=\sqrt{36−9x^2−9y^2}\).
- Hint
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Determine the set of ordered pairs that do not make the radicand negative.
- Solution
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The domain is \(\{(x, y) | x^2+y^2≤4 \}\) the shaded circle defined by the inequality \(x^2+y^2≤4\), which has a circle of radius \(2\) as its boundary. The range is \([0,6].\)
Graphing Functions of Two Variables
Suppose we wish to graph the function \(z=f(x,y).\) This function has two independent variables (\(x\) and \(y\)) and one dependent variable \((z)\). When graphing a function \(y=f(x)\) of one variable, we use the Cartesian plane. We are able to graph any ordered pair \((x,y)\) in the plane, and every point in the plane has an ordered pair \((x,y)\) associated with it. With a function of two variables, each ordered pair \((x,y)\) in the domain of the function is mapped to a real number \(z\). Therefore, the graph of the function \(f\) consists of ordered triples \((x,y,z)\). The graph of a function \(z=f(x,y)\) of two variables is called a surface.
To understand more completely the concept of plotting a set of ordered triples to obtain a surface in three-dimensional space, imagine the \((x,y)\) coordinate system laying flat. Then, every point in the domain of the function f has a unique \(z\)-value associated with it. If \(z\) is positive, then the graphed point is located above the \(xy\)-plane, if \(z\) is negative, then the graphed point is located below the \(xy\)-plane. The set of all the graphed points becomes the two-dimensional surface that is the graph of the function \(f\).
Example \(\PageIndex{2}\): Graphing Functions of Two Variables
Create a graph of each of the following functions:
- \(g(x,y)=\sqrt{9−x^2−y^2}\)
- \(f(x,y)=x^2+y^2\)
Solution
a. In Example \(\PageIndex{2}\), we determined that the domain of \(g(x,y)=\sqrt{9−x^2−y^2}\) is \(\{(x,y)∈R^2∣x^2+y^2≤9\}\) and the range is \(\{z∈R^2∣0≤z≤3\}\). When \(x^2+y^2=9\) we have \(g(x,y)=0\). Therefore any point on the circle of radius \(3\) centered at the origin in the \(xy\)-plane maps to \(z=0\) in \(R^3\). If \(x^2+y^2=8\), then \(g(x,y)=1,\) so any point on the circle of radius \(2\sqrt{2}\) centered at the origin in the \(xy\)-plane maps to \(z=1\) in \(R^3\). As \(x^2+y^2\) gets closer to zero, the value of \(z\) approaches \(3\). When \(x^2+y^2=0\), then \(g(x,y)=3\). This is the origin in the \(xy\)-plane If \(x^2+y^2\) is equal to any other value between \(0\) and \(9\), then \(g(x,y)\) equals some other constant between \(0\) and \(3\). The surface described by this function is a hemisphere centered at the origin with radius \(3\) as shown in the following graph.
b. This function also contains the expression \(x^2+y^2\). Setting this expression equal to various values starting at zero, we obtain circles of increasing radius. The minimum value of \(f(x,y)=x^2+y^2\) is zero (attained when \(x=y=0.\). When \(x=0\), the function becomes \(z=y^2\), and when \(y=0\), then the function becomes \(z=x^2\). These are cross-sections of the graph, and are parabolas. Recall from Introduction to Vectors in Space that the name of the graph of \(f(x,y)=x^2+y^2\) is a paraboloid . The graph of \(f\) appears in the following graph.
Example \(\PageIndex{3}\): Nuts and Bolts
A profit function for a hardware manufacturer is given by
\[f(x,y)=16−(x−3)^2−(y−2)^2, \nonumber\]
where \(x\) is the number of nuts sold per month (measured in thousands) and \(y\) represents the number of bolts sold per month (measured in thousands). Profit is measured in thousands of dollars. Sketch a graph of this function.
Solution
This function is a polynomial function in two variables. The domain of \(f\) consists of \((x,y)\) coordinate pairs that yield a nonnegative profit:
\[ \begin{align*} 16−(x−3)^2−(y−2)^2 &≥ 0 \\[5pt] (x−3)^2+(y−2)^2 &≤ 16. \end{align*}\]
This is a disk of radius \(4\) centered at \((3,2)\). A further restriction is that both \(x\) and \(y\) must be nonnegative. When \(x=3\) and \(y=2, \quad f(x,y)=16.\) Note that it is possible for either value to be a noninteger; for example, it is possible to sell \(2.5\) thousand nuts in a month. The domain, therefore, contains thousands of points, so we can consider all points within the disk. For any \(z<16\), we can solve the equation \(f(x,y)=16:\)
\[ \begin{align*} 16−(x−3)^2−(y−2)^2 &=z \\[5pt] (x−3)^2+(y−2)^2 &=16−z. \end{align*}\]
Since \(z<16,\) we know that \(16−z>0,\) so the previous equation describes a circle with radius \(\sqrt{16−z}\) centered at the point \((3,2)\). Therefore. the range of \(f(x,y)\) is \(\{z∈\mathbb{R}|z≤16\}.\) The graph of \(f(x,y)\) is also a paraboloid, and this paraboloid points downward as shown.
Level Curves
If hikers walk along rugged trails, they might use a topographical map that shows how steeply the trails change. A topographical map contains curved lines called contour lines. Each contour line corresponds to the points on the map that have equal elevation (Figure \(\PageIndex{6}\)). A level curve of a function of two variables \(f(x,y)\) is completely analogous to a contour line on a topographical map.
Definition: Level Curves
Given a function \(f(x,y)\) and a number \(c\) in the range of \(f\), a level curve of a function of two variables for the value \(c\) is defined to be the set of points satisfying the equation \(f(x,y)=c.\)
Returning to the function \(g(x,y)=\sqrt{9−x^2−y^2}\), we can determine the level curves of this function. The range of \(g\) is the closed interval \([0,3]\). First, we choose any number in this closed interval—say, \(c=2\). The level curve corresponding to \(c=2\) is described by the equation
\[ \sqrt{9−x^2−y^2}=2.\]
To simplify, square both sides of this equation:
\[ 9−x^2−y^2=4.\]
Now, multiply both sides of the equation by \(−1\) and add \(9\) to each side:
\[ x^2+y^2=5.\]
This equation describes a circle centered at the origin with radius \(\sqrt{5}\). Using values of \(c\) between \(0\) and \(3\) yields other circles also centered at the origin. If \(c=3\), then the circle has radius \(0\), so it consists solely of the origin. Figure \(\PageIndex{7}\) is a graph of the level curves of this function corresponding to \(c=0,1,2,\) and \(3\). Note that in the previous derivation it may be possible that we introduced extra solutions by squaring both sides. This is not the case here because the range of the square root function is nonnegative.
A graph of the various level curves of a function is called a contour map .
Example \(\PageIndex{4}\): Making a Contour Map
Given the function \(f(x,y)=\sqrt{8+8x−4y−4x^2−y^2}\), find the level curve corresponding to \(c=0\). Then create a contour map for this function. What are the domain and range of \(f\)?
Solution
To find the level curve for \(c=0,\) we set \(f(x,y)=0\) and solve. This gives
\(0=\sqrt{8+8x−4y−4x^2−y^2}\).
We then square both sides and multiply both sides of the equation by \(−1\):
\(4x^2+y^2−8x+4y−8=0.\)
Now, we rearrange the terms, putting the \(x\) terms together and the \(y\) terms together, and add \(8\) to each side:
\(4x^2−8x+y^2+4y=8.\)
Next, we group the pairs of terms containing the same variable in parentheses, and factor \(4\) from the first pair:
\(4(x^2−2x)+(y^2+4y)=8.\)
Then we complete the square in each pair of parentheses and add the correct value to the right-hand side:
\(4(x^2−2x+1)+(y^2+4y+4)=8+4(1)+4.\)
Next, we factor the left-hand side and simplify the right-hand side:
\(4(x−1)^2+(y+2)^2=16.\)
Last, we divide both sides by \(16\):
\[\dfrac{(x−1)^2}{4}+\dfrac{(y+2)^2}{16}=1. \label{ellipseeq1}\]
This equation describes an ellipse centered at \((1,−2)\). The graph of this ellipse appears in the following graph.
We can repeat the same derivation for values of \(c\) less than \(4.\) Then, Equation \ref{ellipseeq1} becomes
\(\dfrac{4(x−1)^2}{16−c^2}+\dfrac{(y+2)^2}{16−c^2}=1\)
for an arbitrary value of \(c\).
Setting the function equal to \(c\), we get:
\[\sqrt{8+8x−4y−4x^2−y^2} = c\nonumber\]
We then square both sides and multiply both sides of the equation by \(−1\):
\(4x^2+y^2−8x+4y−8=-c^2.\)
Now, we rearrange the terms, putting the \(x\) terms together and the \(y\) terms together, and add \(8\) to each side:
\(4x^2−8x+y^2+4y=8-c^2.\)
Next, we group the pairs of terms containing the same variable in parentheses, and factor \(4\) from the first pair:
\(4(x^2−2x)+(y^2+4y)=8-c^2.\)
Then we complete the square in each pair of parentheses and add the correct value to the right-hand side:
\(4(x^2−2x+1)+(y^2+4y+4)=8 - c^2 +4(1)+4.\)
Next, we factor the left-hand side and simplify the right-hand side:
\(4(x−1)^2+(y+2)^2=16 - c^2.\)
Last, we divide both sides by \(16 - c^2\):
\[\dfrac{4(x−1)^2}{16−c^2}+\dfrac{(y+2)^2}{16−c^2}=1 \nonumber\]
Figure \(\PageIndex{9}\) shows a contour map for \(f(x,y)\) using the values \(c=0,1,2,\) and \(3\). When \(c=4,\) the level curve is the point \((−1,2)\).
Finding the Domain & Range
Since this is a square root function, the radicand must not be negative. So we have
\[8+8x−4y−4x^2−y^2\ge 0 \nonumber\]
Recognizing that the boundary of the domain is an ellipse, we repeat the steps we showed above to obtain
\[\dfrac{(x−1)^2}{4}+\dfrac{(y+2)^2}{16}\le 1 \nonumber\]
So the domain of \(f\) can be written: \(\big\{ (x,y) \,|\, \frac{(x−1)^2}{4}+\frac{(y+2)^2}{16}\le 1 \big\}.\)
To find the range of \(f,\) we need to consider the possible outputs of this square root function. We know the output cannot be negative, so we need to next check if its output is ever \(0.\) From the work we completed above to find the level curve for \(c = 0,\) we know the value of \(f\) is \(0\) for any point on that level curve (on the ellipse, \(\frac{(x−1)^2}{4}+\frac{(y+2)^2}{16}=1\)). So we know the lower bound of the range of this function is \(0.\)
To determine the upper bound for the range of the function in this problem, it's easier if we first complete the square under the radical.
\[\begin{align*} f(x,y) &= \sqrt{8+8x−4y−4x^2−y^2} \\[5pt]
&= \sqrt{8 - 4(x^2 - 2x \quad) - (y^2 + 4y \quad)}\\[5pt]
&= \sqrt{8 - 4(x^2 - 2x +1 - 1) - (y^2 + 4y + 4 - 4)}\\[5pt]
&= \sqrt{8 - 4(x^2 - 2x +1) + 4 - (y^2 + 4y + 4) +4}\\[5pt]
&= \sqrt{16 - 4(x-1)^2 - (y+2)^2} \end{align*}\]
Now that we have \(f\) in this form, we can see how large the radicand can be. Since we are subtracting two perfect squares from \(16,\) we know that the value of the radicand cannot be greater than \(16.\) At the point \((1, -2),\) we can see the radicand will be 16 (since we will be subtracting \(0\) from \(16\) at this point. This gives us the maximum value of \(f\), that is \(f(1, -2) = \sqrt{16} = 4.\)
So the range of this function is \([0, 4].\)
Exercise \(\PageIndex{2}\)
Find and graph the level curve of the function \(g(x,y)=x^2+y^2−6x+2y\) corresponding to \(c=15.\)
- Hint
-
First, set \(g(x,y)=15\) and then complete the square.
- Solution
-
The equation of the level curve can be written as \((x−3)^2+(y+1)^2=25,\) which is a circle with radius \(5\) centered at \((3,−1).\)
Another useful tool for understanding the graph of a function of two variables is called a vertical trace. Level curves are always graphed in the \(xy\)-plane, but as their name implies, vertical traces are graphed in the \(xz-\) or \(yz-\) planes.
Definition: Vertical Traces
Consider a function \(z=f(x,y)\) with domain \(D⊆\rm I\!R^2\). A vertical trace of the function can be either the set of points that solves the equation \(f(a,y)=z\) for a given constant \(x=a\) or \(f(x,b)=z\) for a given constant \(y=b.\)
Example \(\PageIndex{5}\): Finding Vertical Traces
Find vertical traces for the function \(f(x,y)=\sin x \cos y\) corresponding to \(x=−\dfrac{π}{4},0,\) and \(\dfrac{π}{4}\), and \(y=−\dfrac{π}{4},0\), and \(\dfrac{π}{4}\).
Solution
First set \(x=−\dfrac{π}{4}\) in the equation \(z=\sin x \cos y:\)
\(z=\sin(−\dfrac{π}{4})\cos y=−\dfrac{\sqrt{2}\cos y}{2}≈−0.7071\cos y.\)
This describes a cosine graph in the plane \(x=−\dfrac{π}{4}\). The other values of \(z\) appear in the following table.
| \(c\) | Vertical Trace for \(x=c\) |
|---|---|
| \(−\dfrac{π}{4}\) | \(z=−\dfrac{\sqrt{2}\cos y}{2}\) |
| 0 | \(z=0\) |
| \(\dfrac{π}{4}\) | \(z=\dfrac{\sqrt{2}\cos y}{2}\) |
In a similar fashion, we can substitute the \(y\)-values in the equation \(f(x,y)\) to obtain the traces in the \(yz\)-plane, as listed in the following table.
| \(d\) | Vertical Trace for \(y=d\) |
|---|---|
| \(\dfrac{π}{4}\) | \(z=\dfrac{\sqrt{2}\sin x}{2}\) |
| 0 | \(z=\sin x\) |
| \(−\dfrac{π}{4}\) | \(z=\dfrac{\sqrt{2}\sin x}{2}\) |
The three traces in the \(xz\)-plane are cosine functions; the three traces in the \(yz\)-plane are sine functions. These curves appear in the intersections of the surface with the planes \(x=−\dfrac{π}{4},\quad x=0,\quad x=\dfrac{π}{4}\) and \(y=−\dfrac{π}{4},\quad y=0,\quad y=\dfrac{π}{4}\) as shown in the following figure.
Exercise \(\PageIndex{3}\)
Determine the equation of the vertical trace of the function \(g(x,y)=−x^2−y^2+2x+4y−1\) corresponding to \(y=3\), and describe its graph.
- Hint
-
Set \(y=3\) in the equation \(z=−x^2−y^2+2x+4y−1\) and complete the square.
- Solution
-
\(z=3−(x−1)^2\). This function describes a parabola opening downward in the plane \(y=3\).
Functions of two variables can produce some striking-looking surfaces. Figure \(\PageIndex{11}\) shows two examples.
Functions of More Than Two Variables
So far, we have examined only functions of two variables. However, it is useful to take a brief look at functions of more than two variables. Two such examples are
\[ \underbrace{f(x,y,z)=x^2−2xy+y^2+3yz−z^2+4x−2y+3x−6}_{\text{a polynomial in three variables}}\]
and
\[g(x,y,t)=(x^2−4xy+y^2)\sin t−(3x+5y)\cos t.\]
In the first function, \((x,y,z)\) represents a point in space, and the function \(f\) maps each point in space to a fourth quantity, such as temperature or wind speed. In the second function, \((x,y)\) can represent a point in the plane, and \(t\) can represent time. The function might map a point in the plane to a third quantity (for example, pressure) at a given time \(t\). The method for finding the domain of a function of more than two variables is analogous to the method for functions of one or two variables.
Example \(\PageIndex{6}\): Domains for Functions of Three Variables
Find the domain of each of the following functions:
- \(f(x,y,z)=\dfrac{3x−4y+2z}{\sqrt{9−x^2−y^2−z^2}}\)
- \(g(x,y,t)=\dfrac{\sqrt{2t−4}}{x^2−y^2}\)
Solution :
a. For the function \(f(x,y,z)=\dfrac{3x−4y+2z}{\sqrt{9−x^2−y^2−z^2}}\) to be defined (and be a real value), two conditions must hold:
- The denominator cannot be zero.
- The radicand cannot be negative.
Combining these conditions leads to the inequality
\[9−x^2−y^2−z^2>0.\nonumber\]
Moving the variables to the other side and reversing the inequality gives the domain as
\[domain(f)=\{(x,y,z)∈R^3∣x^2+y^2+z^2<9\},\nonumber\]
which describes a ball of radius \(3\) centered at the origin. (Note: The surface of the ball is not included in this domain.)
b. For the function \(g(x,y,t)=\dfrac{\sqrt{2t−4}}{x^2−y^2}\) to be defined (and be a real value), two conditions must hold:
- The radicand cannot be negative.
- The denominator cannot be zero.
Since the radicand cannot be negative, this implies \(2t−4≥0\), and therefore that \(t≥2\). Since the denominator cannot be zero, \(x^2−y^2≠0\), or \(x^2≠y^2\), Which can be rewritten as \(y=±x\), which are the equations of two lines passing through the origin. Therefore, the domain of \(g\) is
\[ domain(g)=\{(x,y,t)|y≠±x,t≥2\}. \nonumber\]
Exercise \(\PageIndex{4}\)
Find the domain of the function \(h(x,y,t)=(3t−6)\sqrt{y−4x^2+4}\).
- Hint
-
Check for values that make radicands negative or denominators equal to zero.
- Solution
-
\[domain(h)=\{(x,y,t)\in \mathbb{R}^3∣y≥4x^2−4\} \nonumber\]
Functions of two variables have level curves, which are shown as curves in the \(xy-plane.\) However, when the function has three variables, the curves become surfaces, so we can define level surfaces for functions of three variables.
Definition: Level Surface of a Function of Three Variables
Given a function \(f(x,y,z)\) and a number \(c\) in the range of \(f\), a level surface of a function of three variables is defined to be the set of points satisfying the equation \(f(x,y,z)=c.\)
Example \(\PageIndex{7}\): Finding a Level Surface
Find the level surface for the function \(f(x,y,z)=4x^2+9y^2−z^2\) corresponding to \(c=1\).
Solution
The level surface is defined by the equation \(4x^2+9y^2−z^2=1.\) This equation describes a hyperboloid of one sheet as shown in Figure \(\PageIndex{12}\).
Exercise \(\PageIndex{5}\)
Find the equation of the level surface of the function
\[ g(x,y,z)=x^2+y^2+z^2−2x+4y−6z \nonumber\]
corresponding to \(c=2,\) and describe the surface, if possible.
- Hint
-
Set \(g(x,y,z)=c\) and complete the square.
- Solution
-
\((x−1)^2+(y+2)^2+(z−3)^2=16\) describes a sphere of radius \(4\) centered at the point \((1,−2,3).\)
Summary
- The graph of a function of two variables is a surface in \(\mathbb{R}^3\) and can be studied using level curves and vertical traces.
- A set of level curves is called a contour map.
Key Equations
- Vertical trace
\(f(a,y)=z\) for \(x=a\) or \(f(x,b)=z\) for \(y=b\)
- Level surface of a function of three variables
\(f(x,y,z)=c\)
Glossary
- contour map
- a plot of the various level curves of a given function \(f(x,y)\)
- function of two variables
- a function \(z=f(x,y)\) that maps each ordered pair \((x,y)\) in a subset \(D\) of \(R^2\) to a unique real number \(z\)
- graph of a function of two variables
- a set of ordered triples \((x,y,z)\) that satisfies the equation \(z=f(x,y)\) plotted in three-dimensional Cartesian space
- level curve of a function of two variables
- the set of points satisfying the equation \(f(x,y)=c\) for some real number \(c\) in the range of \(f\)
- level surface of a function of three variables
- the set of points satisfying the equation \(f(x,y,z)=c\) for some real number \(c\) in the range of \(f\)
- surface
- the graph of a function of two variables, \(z=f(x,y)\)
- vertical trace
- the set of ordered triples \((c,y,z)\) that solves the equation \(f(c,y)=z\) for a given constant \(x=c\) or the set of ordered triples \((x,d,z)\) that solves the equation \(f(x,d)=z\) for a given constant \(y=d\)
Contributors
-
Gilbert Strang (MIT) and Edwin “Jed” Herman (Harvey Mudd) with many contributing authors. This content by OpenStax is licensed with a CC-BY-SA-NC 4.0 license. Download for free at http://cnx.org .


2.6: Graphing Points on a Coordinate Plane
Learning outcome
- Graph points (ordered pairs) on a coordinate plane
Many maps, such as the Campus Map shown below, use a grid system to identify locations. Do you see the numbers \(1,2,3\), and \(4\) across the top and bottom of the map and the letters A, B, C, and D along the sides? Every location on the map can be identified by a number and a letter.
For example, the Student Center is in section 2B. It is located in the grid section above the number \(2\) and next to the letter B. In which grid section is the Stadium? The Stadium is in section 4D.
example
Use the map above.
- Find the grid section of the Residence Halls.
- What is located in grid section 4C?
Solution
- Read the number below the Residence Halls, \(4\), and the letter to the side, A. So the Residence Halls are in grid section 4A.
- Find \(4\) across the bottom of the map and C along the side. Look below the \(4\) and next to the C. Tiger Field is in grid section 4C.
try it
[ohm_question]146887[/ohm_question]
[ohm_question]146888[/ohm_question]
Just as maps use a grid system to identify locations, a grid system is used in algebra to show a relationship between two variables in a rectangular coordinate system. To create a rectangular coordinate system, start with a horizontal number line. Show both positive and negative numbers as you did before, using a convenient scale unit. This horizontal number line is called the \(x\)-axis.
Now, make a vertical number line passing through the \(x\)-axis at \(0\). Put the positive numbers above \(0\) and the negative numbers below \(0\). See the image below. This vertical line is called the\(y\)-axis.
Vertical grid lines pass through the integers marked on the \(x\)-axis. Horizontal grid lines pass through the integers marked on the \(y\)-axis]. The resulting grid is the rectangular coordinate system.
The rectangular coordinate system is also called the \(x\text{-}y\) plane, the coordinate plane, or the Cartesian coordinate system (since it was developed by a mathematician named René Descartes.)
The rectangular coordinate system
The \(x\)-axis] and the \(y\)-axis form the rectangular coordinate system. These axes divide a plane into four areas, called quadrants. The quadrants are identified by Roman numerals, beginning on the upper right and proceeding counterclockwise. See the image below.
The four quadrants of the rectangular coordinate system
In the rectangular coordinate system, every point is represented by an ordered pair. The first number in the ordered pair is the x -coordinate of the point, and the second number is the y -coordinate of the point.
Ordered Pair
An ordered pair, \(\left(x,y\right)\) gives the coordinates of a point in a rectangular coordinate system.
\(\begin{array}{c}\text{The first number is the }x\text{-coordinate}.\hfill \\ \text{The second number is the }y\text{-coordinate}.\hfill \end{array}\)
So how do the coordinates of a point help you locate a point on the \(x\text{-}y\) plane?
Let’s try locating the point \(\left(2,5\right)\). In this ordered pair, the \(x\) -coordinate is \(2\) and the \(y\) -coordinate is \(5\).
We start by locating the \(x\) value, \(2\), on the \(x\text{-axis.}\) Then we lightly sketch a vertical line through \(x=2\), as shown in the image below.
Now we locate the \(y\) value, \(5\), on the \(y\) -axis and sketch a horizontal line through \(y=5\). The point where these two lines meet is the point with coordinates \(\left(2,5\right)\). We plot the point there, as shown in the image below.
example
Plot \(\left(1,3\right)\) and \(\left(3,1\right)\) in the same rectangular coordinate system.
[reveal-answer q=”501893″]Show Answer[/reveal-answer]
[hidden-answer a=”501893″]
Solution
The coordinate values are the same for both points, but the \(x\) and \(y\) values are reversed. Let’s begin with point \(\left(1,3\right)\). The \(x\text{-coordinate}\) is \(1\) so find \(1\) on the \(x\text{-axis}\) and sketch a vertical line through \(x=1\). The \(y\text{-coordinate}\) is \(3\) so we find \(3\) on the \(y\text{-axis}\) and sketch a horizontal line through \(y=3\). Where the two lines meet, we plot the point \(\left(1,3\right)\).
To plot the point \(\left(3,1\right)\), we start by locating \(3\) on the \(x\text{-axis}\) and sketch a vertical line through \(x=3\). Then we find \(1\) on the \(y\text{-axis}\) and sketch a horizontal line through \(y=1\). Where the two lines meet, we plot the point \(\left(3,1\right)\).
Notice that the order of the coordinates does matter, so, \(\left(1,3\right)\) is not the same point as \(\left(3,1\right)\).
[/hidden-answer]
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[ohm_question]146882[/ohm_question]
Exercises
Plot each point in the rectangular coordinate system and identify the quadrant in which the point is located:
1. \(\left(-1,3\right)\)
2. \(\left(-3,-4\right)\)
3. \(\left(2,-3\right)\)
4. \(\left(3,{\Large\frac{5}{2}}\right)\)
[reveal-answer q=”933080″]Show Answer[/reveal-answer]
[hidden-answer a=”933080″]
Solution
The first number of the coordinate pair is the \(x\text{-coordinate}\), and the second number is the \(y\text{-coordinate}\).
1. Since \(x=-1,y=3\), the point \(\left(-1,3\right)\) is in Quadrant II.
2. Since \(x=-3,y=-4\), the point \(\left(-3,-4\right)\) is in Quadrant III.
3. Since \(x=2,y=-1\), the point \(\left(2,-1\right)\) is in Quadrant lV.
4. Since \(x=3,y={\Large\frac{5}{2}}\), the point \(\left(3,{\Large\frac{5}{2}}\right)\) is in Quadrant I. It may be helpful to write \({\Large\frac{5}{2}}\) as the mixed number, \(2{\Large\frac{1}{2}}\), or decimal, \(2.5\). Then we know that the point is halfway between \(2\) and \(3\) on the \(y\text{-axis}\).
[/hidden-answer]
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[ohm_question]146883[/ohm_question]
example
How do the signs affect the location of the points?
Plot each point:
1. \(\left(-5,2\right)\)
2. \(\left(-5,-2\right)\)
3. \(\left(5,2\right)\)
4. \(\left(5,-2\right)\)
[reveal-answer q=”83169″]Show Answer[/reveal-answer]
[hidden-answer a=”83169″]
Solution
As we locate the \(x\text{-coordinate}\) and the \(y\text{-coordinate}\), we must be careful with the signs.
[/hidden-answer]
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[ohm_question]146885[/ohm_question]
You may have noticed some patterns as you graphed the points in the two previous examples. We can summarize sign patterns of the quadrants as follows. Also see the graph below.
| Quadrant I | Quadrant II | Quadrant III | Quadrant IV |
|---|---|---|---|
| ( x , y ) | ( x , y ) | ( x , y ) | ( x , y ) |
| (+,+) | (−,+) | (−,−) | (+,−) |
What if one coordinate is zero? Where is the point \(\left(0,4\right)\) located? Where is the point \(\left(-2,0\right)\) located? The point \(\left(0,4\right)\) is on the \(y\)-axis and the point \(\left(-2,0\right)\) is on the \(x\)-axis.
Points on the Axes
Points with a \(y\text{-coordinate}\) equal to \(0\) are on the \(x\)-axis, and have coordinates \(\left(a,0\right)\).
Points with an \(x\text{-coordinate}\) equal to \(0\) are on the \(y\)-axis, and have coordinates \(\left(0,b\right)\).
What is the ordered pair of the point where the axes cross? At that point both coordinates are zero, so its ordered pair is \(\left(0,0\right)\). The point has a special name. It is called the origin .
The Origin
The point \(\left(0,0\right)\) is called the origin . It is the point where the \(x\)-axis and \(y\)-axis intersect.
ExAMPLE
Plot each point on a coordinate grid:
1. \(\left(0,5\right)\)
2. \(\left(4,0\right)\)
3. \(\left(-3,0\right)\)
4. \(\left(0,0\right)\)
5. \(\left(0,-1\right)\)
[reveal-answer q=”803167″]Show Answer[/reveal-answer]
[hidden-answer a=”803167″]
Solution
- Since \(x=0\), the point whose coordinates are \(\left(0,5\right)\) is on the \(y\text{-axis}\).
- Since \(y=0\), the point whose coordinates are \(\left(4,0\right)\) is on the \(x\text{-axis}\).
- Since \(y=0\), the point whose coordinates are \(\left(-3,0\right)\) is on the \(x\text{-axis}\).
- Since \(x=0\) and \(y=0\), the point whose coordinates are \(\left(0,0\right)\) is the origin.
-
Since \(x=0\), the point whose coordinates are \(\left(0,-1\right)\) is on the \(y\text{-axis}\).
[/hidden-answer]
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[ohm_question]146886[/ohm_question]
In the following video, we show another example of how to plot several different points on the coordinate plane.
In data and graph analysis, being able to identify the coordinates of a point shown on a graph is just as important as being able to plot points. To identify the \(x\)-coordinate of a point on a graph, read the number on the \(x\)-axis directly above or below the point. To identify the \(y\)-coordinate of a point, read the number on the \(y\)-axis directly to the left or right of the point. Remember, to write the ordered pair using the correct order \(\left(x,y\right)\).
example
Name the ordered pair of each point shown:
Solution
Point A is above \(-3\) on the \(x\text{-axis}\), so the \(x\text{-coordinate}\) of the point is \(-3\). The point is to the left of \(3\) on the \(y\text{-axis}\), so the \(y\text{-coordinate}\) of the point is \(3\). The coordinates of the point are \(\left(-3,3\right)\).
Point B is below \(-1\) on the \(x\text{-axis}\), so the \(x\text{-coordinate}\) of the point is \(-1\). The point is to the left of \(-3\) on the \(y\text{-axis}\), so the \(y\text{-coordinate}\) of the point is \(-3\). The coordinates of the point are \(\left(-1,-3\right)\).
Point C is above \(2\) on the \(x\text{-axis}\), so the \(x\text{-coordinate}\) of the point is \(2\). The point is to the right of \(4\) on the \(y\text{-axis}\), so the \(y\text{-coordinate}\) of the point is \(4\). The coordinates of the point are \(\left(2,4\right)\).
Point D is below \(4\) on the \(x-\text{axis}\), so the \(x\text{-coordinate}\) of the point is \(4\). The point is to the right of \(-4\) on the \(y\text{-axis}\), so the \(y\text{-coordinate}\) of the point is \(-4\). The coordinates of the point are \(\left(4,-4\right)\).
try it
[ohm_question]146915[/ohm_question]
example
Name the ordered pair of each point shown:
[reveal-answer q=”228616″]Show Answer[/reveal-answer]
[hidden-answer a=”228616″]
Solution
| Point A is on the x -axis at \(x=-4\) | The coordinates of point A are \(\left(-4,0\right)\) |
| Point B is on the y -axis at \(y=-2\) | The coordinates of point B are \(\left(0,-2\right)\) |
| Point C is on the x -axis at \(x=3\) | The coordinates of point C are \(\left(3,0\right)\) |
| Point D is on the y- axis at \(y=1\) | The coordinates of point D are \(\left(0,1\right)\) |
[/hidden-answer]
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[ohm_question]146919[/ohm_question]
Watch the following video for another example of how to determine the ordered pair for points on the coordinate plane.


9.1.4: Acid-base reactions
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Acid-base reactions are chemical reactions that involve the transfer of a proton (H + ). Examples include the reactions of acids with metals, carbonates, and Arrhenius bases, described in the following.
Reactions of acids with metals
Metals tend to give out an electron and become cations. The majority of the metals, called reactive metals, give out electrons to protons in the acids and release H 2 gas. For example, Fig. 6.7.1 shows magnesium reacting with hydrochloric acid by the following reaction:
\[\mathrm{Mg}(\mathrm{s})+2 \mathrm{HCl} \rightarrow \mathrm{MgCl}_{2}(\mathrm{aq})+\mathrm{H}_{2}(\mathrm{~g}) \uparrow\nonumber\]
Alkali metals like Na give out one electron, alkaline earth metals like Mg give out two electrons, and aluminum gives out three electrons. A proton can accept one electron. Therefore, the number of acidic protons should be equal to the number of electrons given out to balance the electron transfer in these reactions.
Note
Noble metals or jewelry metals like gold, silver, platinum, copper are exceptions –they do not react with acids.
Example \(\PageIndex{1}\)
write a balanced equation for the reaction of aluminum with HBr?
Solution
Step 1) write the equation with the correct formulas of reactants and products.
\[\mathrm{Al}(\mathrm{s})+\mathrm{HBr}(\mathrm{aq}) \rightarrow \mathrm{AlBr}_{3}(\mathrm{aq})+\mathrm{H}_{2}(\mathrm{~g})\nonumber\]
Note that Al gives out three electrons to make Al 3 + ; that is why three Br - are attached to it in the product to balance the charge.
Step 2) balance the electrons lost by the metal with the number of acidic protons.
\[\mathrm{Al}(\mathrm{s})+3 \mathrm{HBr}(\mathrm{aq}) \rightarrow \mathrm{AlBr}_{3}(\mathrm{aq})+\mathrm{H}_{2}(\mathrm{~g})\nonumber\]
Step 3) Balance the rest of the elements by the hit and trial method. There are 3 hydrogen atoms on the left, balance them by adding 3/2 coefficient to the hydrogen on the right.
\[\mathrm{Al}(\mathrm{s})+3 \mathrm{HBr}(\mathrm{aq}) \rightarrow \mathrm{AlBr}_{3}(\mathrm{aq})+\frac{3}{2} \mathrm{H}_{2}(\mathrm{~g})\nonumber\]
Step 4. Although the equation is balanced, it is recommended to remove the fraction by multiplying the coefficients in the whole equation with the highest common factor to obtain the final balanced equation.
\[2 \mathrm{Al}(\mathrm{s})+6 \mathrm{HBr}(\mathrm{aq}) \rightarrow 2 \mathrm{AlBr}_{3}(\mathrm{aq})+3 \mathrm{H}_{2}(\mathrm{~g})\nonumber\]
Reactions of acids with carbonates and hydrogen carbonates
Carbonic acid (H 2 CO 3 ) is a weak acid found in carbonated water. The H 2 CO 3 is a product of carbon dioxide (CO 2 ) and water (H 2 O) by the following equilibrium reaction.
\[\ce{CO2(aq) + H2O(l) <=> H2CO3(aq)}\nonumber\]
Hydrogen carbonate (HCO 3 - ) and carbonate (CO 3 2 - ) are one and two hydrogen less, respectively than carbonic acid. The salts containing HCO 3 - and CO 3 2 - accept one and two protons, respectively, from acids to make H 2 CO 3 . The H 2 CO 3 decomposes to carbon dioxide and water by the reverse reaction shown above. Fig. 6.7.2 shows the reaction of sodium hydrogen carbonate with hydrochloric acid.
\[\mathrm{NaHCO}_{3}(\mathrm{~s})+\mathrm{HCl}(\mathrm{aq}) \rightarrow \mathrm{NaCl}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\mathrm{I})+\mathrm{CO}_{2}(\mathrm{~g}) \uparrow\nonumber\]
Similarly, calcium carbonate found in limestone reacts with acids. For example, sulfuric acid is one of the components in acid rain that reacts with calcium carbonate and damages sculptures made of stone.
\[\mathrm{CaCO}_{3}(\mathrm{~s})+\mathrm{H}_{2} \mathrm{SO}_{4}(\mathrm{aq}) \rightarrow \mathrm{CaSO}_{4}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l})+\mathrm{CO}_{2}(\mathrm{~g}) \uparrow\nonumber\]
Reactions of acids with Arrhenius bases
Acids release proton (H + ) and Arrhenius bases release hydroxide ions (OH - ) in solution. When an acid mix with the Arrhenius base, H + and OH - ions react with each other and produce water molecules.
\[\mathrm{H}^{+}(\mathrm{aq})+\mathrm{OH}^{-}(\mathrm{aq}) \rightarrow \mathrm{H}_{2} \mathrm{O}(\mathrm{l})\nonumber\]
This is the net ionic equation of a reaction of an acid with an Arrhenius base. The cation of the base and the anion of the acid do not react –they are spectator ions. The reaction equation for an acid-base reaction is written in various ways explained below.
Molecular equation
The molecular equation shows the formulas of the substances. For example, the molecular equation of the reaction of HCl with NaOH is the following.
\[\mathrm{HCl}(\mathrm{aq})+\mathrm{NaOH}(a q) \rightarrow \mathrm{NaCl}(a q)+\mathrm{H}_{2} \mathrm{O}(\mathrm{l})\nonumber\]
Complete ionic equation
The complete ionic equation shows the electrolytes dissociated into ions as they actually exist in the water phase, i.e., as aqueous ions. The complete ionic equation of the reaction of HCl with NaOH is the following.
\[\mathrm{H}^{+}(\mathrm{aq})+\mathrm{Cl}^{-}(\mathrm{aq})+\mathrm{Na}^{+}(\mathrm{aq})+\mathrm{OH}^{-}(\mathrm{aq}) \rightarrow \mathrm{H}_{2} \mathrm{O}(\mathrm{l})+\mathrm{Na}^{+}(\mathrm{aq})+\mathrm{Cl}^{-}(\mathrm{aq})\nonumber\]
It is evident that Na + and Cl - do not react –they are the spectator ions.
Net ionic equation
The spectator ions are present on both sides of the equation in equal numbers. The spectator ions can be canceled out, like the terms in an algebraic equation.
\[\mathrm{H}^{+}(\mathrm{aq})+\cancel{\mathrm{Cl}^{-}(\mathrm{aq})}+\cancel{\mathrm{Na}^{+}(\mathrm{aq})}+\mathrm{OH}^{-}(\mathrm{aq}) \rightarrow \mathrm{H}_{2} \mathrm{O}(\mathrm{l})+\cancel{\mathrm{Na}^{+}(\mathrm{aq})}+\cancel{\mathrm{Cl}^{-}(\mathrm{aq})}\nonumber\]
The net ionic equation shows the substances that are not spectator ions after canceling out the spectator ions from the complete ionic equation. The net ionic equation of the reaction of HCl with NaOH is the following.
\[\mathrm{H}^{+}(\mathrm{aq})+\mathrm{OH}^{-}(\mathrm{aq}) \rightarrow \mathrm{H}_{2} \mathrm{O}(\mathrm{l})\nonumber\]
Note
All of the reactions of the strong acids with strong Arrhenius bases have the same net ionic equation.
Writing a balanced chemical equation of the reaction of acids with Arrhenius bases
The following example shows the steps needed to write the balanced equation of the reaction of an acid with an Arrhenius base.
Example \(\PageIndex{2}\)
Write a balanced chemical equation of a reaction between HCl and Ca(OH) 2 ?
Solution
Step 1) Write the formula of acid and base in the reactants and salt and water in the products. All the strong electrolytes dissolve in water, so use (aq) to represent their state.
\[\mathrm{HCl}(\mathrm{aq})+\mathrm{Ca}(\mathrm{OH})_{2}(\mathrm{aq}) \rightarrow \text { Salt(aq) }+\mathrm{H}_{2} \mathrm{O}(\mathrm{l})\nonumber\]
Step 2) Balance the H + in the acid with the OH - in the base.
\[2 \mathrm{HCl}(\mathrm{aq})+\mathrm{Ca}(\mathrm{OH})_{2}(\mathrm{aq}) \rightarrow \mathrm{Salt}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l})\nonumber\]
Step 3) Balance the H 2 O with the H + in the acid, or with the OH - in the base.
\[2 \mathrm{HCl}(\mathrm{aq})+\mathrm{Ca}(\mathrm{OH})_{2}(\mathrm{aq}) \rightarrow \text { Salt }(\mathrm{aq})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{I})\nonumber\]
Step 4) Write the salt by combing the cations from the base and the anions from the acid. Make sure the charges are balanced in the salt to make it a neutral substance.
\[2 \mathrm{HCl}(\mathrm{aq})+\mathrm{Ca}(\mathrm{OH})_{2}(\mathrm{aq}) \rightarrow \mathrm{CaCl}_{2}(\mathrm{aq})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{I})\]
Example \(\PageIndex{3}\)
Write a balanced chemical equation of a reaction between H 2 SO 4 and Ba(OH) 2 ?
Solution
Step 1) Write the formula of acid and base in the reactants and salt and water in the products. All the strong electrolytes dissolve in water, so use (aq) to represent their state.
\[\mathrm{H}_{2} \mathrm{SO}_{4}(\mathrm{aq})+\mathrm{Ba}(\mathrm{OH})_{2}(\mathrm{aq}) \rightarrow \mathrm{Salt}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l})\nonumber\]
Step 2) Balance the H + in the acid with the OH - in the base. They are already balanced in the above equation.
Step 3) Balance the H 2 O with the H + in the acid, or with the OH - in the base.
\[\mathrm{H}_{2} \mathrm{SO}_{4}(\mathrm{aq})+\mathrm{Ba}(\mathrm{OH})_{2}(\mathrm{aq}) \rightarrow \mathrm{Salt}(\mathrm{aq})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l})\nonumber\]
Step 4) Write the salt by combing the cations from the base and the anions from the acid. Make sure the charges are balanced in the salt to make it a neutral substance.
\[\mathrm{H}_{2} \mathrm{SO}_{4}(\mathrm{aq})+\mathrm{Ba}(\mathrm{OH})_{2}(\mathrm{aq}) \rightarrow \mathrm{BaSO}_{4}(\mathrm{aq})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l})\nonumber\]
Antacids
The stomach sometimes produces excess HCl that may cause heartburn.
Antacids are the substances used to neutralize the excess HCl in the stomach.
Fig. 6.7.3 shows antacid tablets. The antacids include Arrhenius bases with very low solubility in water, like Al(OH) 3 , and Mg(OH) 2, or weak bases like CaCO 2 , and NaHCO 3 . that react with and neutralize the acid:
\[\begin{aligned}
&\mathrm{Al}(\mathrm{OH})_{3}(\mathrm{~s})+3 \mathrm{HCl}(\mathrm{aq}) \rightarrow \mathrm{AlCl}_{3}(\mathrm{aq})+3 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \\
&\mathrm{Mg}(\mathrm{OH})_{2}(\mathrm{~s})+2 \mathrm{HCl}(\mathrm{aq}) \rightarrow \mathrm{MgCl}_{2}(\mathrm{aq})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \\
&\mathrm{CaCO}_{3}(\mathrm{~s})+2 \mathrm{HCl}(\mathrm{aq}) \rightarrow \mathrm{CaCl}_{2}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\mathrm{I})+\mathrm{CO}_{2}(\mathrm{~g}) \\
&\mathrm{NaHCO}_{3}(\mathrm{~s})+\mathrm{HCl}(\mathrm{aq}) \rightarrow \mathrm{NaCl}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\mathrm{I})+\mathrm{CO}_{2}(g)
\end{aligned}\nonumber\]
Side effects of the antacids
The antacids have side effects, e.g., aluminum and calcium-containing antacids may cause constipation. Magnesium-containing antacids have a laxative effect. Some formulations use a mixture of aluminum and magnesium-containing antacids to cancel out the side effect of each other. The calcium-containing antacid is not recommended for persons who tend to develop kidney stones, because the kidney stone is usually a salt of calcium. Sodium-containing antacids may increase the sodium level in the body fluid and may raise the blood pH.
Acid-base titration
The acid-base titration is an analytical process of determining the concentration of acid, called analyte , by neutralizing it with a base of know concentration, called the standard , or vice versa.
Usually, the base is in a burette and added drop by drop to a known volume of the acid in an Erlenmeyer flask, as illustrated in Fig. 6.7.4. A few drops of an acid-base indicator are mixed with the acid. The indicator changes color at a specific narrow pH range.
- The point when the stoichiometric amount of the base has been added to the acid is called the equivalence point.
- The point when the indicator changes color is the end-point .
Usually, the equivalence point is almost the same as the end-point.
The calculation steps after the titration are usually the following.
Steps in acid-base titration calculations
- The amount, in moles, of the standard is calculated by multiplying the amount in liters with the molarity of the standard.
- Then, the amount, in moles, of the analyte, is calculated by using the mol-to-mol conversion factor from the balanced chemical equation of the reaction.
- Finally, the molarity of the analyte is calculated by dividing moles by the volume in liters of the analyte.
Example \(\PageIndex{4}\)
What is the molarity of the HCl solution if 50.0 mL of the HCl solution requires 42.0 mL of 0.123 M NaOH solution in the titration?
Solution
Step 1) Given: \[[\mathrm{NaOH}]=0.123 \mathrm{~M}=\frac{0.123 \operatorname{~mol} ~N a O H}{1 \mathrm{~NaOH}}\nonumber\]
\[\text{Volume of standard}=42.0 \mathrm{~mL} .=42.0 \mathrm{~mL} \mathrm{~NaOH} \times \frac{1 \mathrm{~L~NaOH}}{1000 \mathrm{~mL} \mathrm{~NaOH}}=0.0420 ~L \mathrm{~NaOH}\nonumber\]
\[\text{Vol. analyte}=50.0 \mathrm{~mL}=50.0 \mathrm{~mL} \mathrm{~HCl} \times \frac{1 \mathrm{~L~HCl}}{1000 \mathrm{~mL} \mathrm{~HCl}}=0.0500 \mathrm{~L~HCl}\nonumber\]
Step 2) Calculate the moles of the standard from the volume and the molarity product.
\begin{equation}
\text { Moles of } \mathrm{~NaOH} \text {~consumed }=0.0420 \mathrm{~L} \mathrm{~NaOH} \times \frac{0.123 \mathrm{~mol} \mathrm{~NaOH}}{1 \mathrm{~L} \mathrm{~NaOH}}=0.00517 \mathrm{~mol}{ } \mathrm{~NaOH}\nonumber
\end{equation}
Step 3) write the balanced chemical equation of the reaction.
\begin{equation}
\mathrm{HCl}(\mathrm{aq})+\mathrm{NaOH}(\mathrm{aq}) \rightarrow \mathrm{NaCl}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l})\nonumber\
\end{equation}
Step 4) write the conversion factor for mole of standard to mole of analyte calculation from the equation.
\begin{equation}
\frac{1 \text {mol } \mathrm{~HCl}}{1 \mathrm{~mol } \mathrm{~NaOH}}\nonumber
\end{equation}
Step 5) Calculate the moles of the analyte by multiplying the moles of the standard and the conversion factor.
\begin{equation}
0.00517 \mathrm{~mol} \mathrm{~NaOH} \times \frac{1 \mathrm{~mol} \mathrm{~HCl}}{1 \mathrm{~mol} \mathrm{~NaOH}}=0.00517 \mathrm{~mol} { }\mathrm{~HCl}\nonumber
\end{equation}
Step 6) Calculate the molarity of the analyte by dividing the moles with the volume in liters of the analyte.
\begin{equation}
0.00517 \mathrm{~mol} \mathrm{~NaOH} \times \frac{1 \mathrm{~mol} \mathrm{~HCl}}{1 \mathrm{~mol} \mathrm{~NaOH}}=0.00517 \mathrm{~mol}{ } \mathrm{~HCl}\nonumber
\end{equation}
After learning the steps thoroughly, the calculations can be done in a few steps, as shown in the following.
Calculations: \[\quad 42.0 \mathrm{~mL} \mathrm{~NaOH} \times \frac{1 \mathrm{~L} \mathrm{~NaOH}}{1000 \mathrm{~mL} \mathrm{~NaOH}} \times \frac{0.123 \mathrm{~mol} \mathrm{~NaOH}}{1 \mathrm{~L} \mathrm{~NaOH}} \times \frac{1 \mathrm{~mol} \mathrm{~HCl}}{1 \mathrm{~mol} \mathrm{~NaOH}}=0.00517 \mathrm{~mol}{ } \mathrm{~HCl}\nonumber\]
Molarity of HCl: \[\quad M=\frac{n(\mathrm{mol})}{V(L)}=\frac{0.00517 \mathrm{~mol} \mathrm{~HCl}}{50.0 \mathrm{~mL} \mathrm{~HCl}} \times \frac{1000 \mathrm{~mL} \mathrm{~HCl}}{1 \mathrm{~L~HCl}}=0.103 \mathrm{~M} { }\mathrm{~HCl}\nonumber\]
Example \(\PageIndex{5}\)
What is the molarity of the H 2 SO 4 solution if 50.0 mL of the H 2 SO 4 solution requires 32.3 mL of 0.201 M NaOH solution in the titration?
Solution
Step 1) Given: \begin{equation}
\begin{aligned}
&{[\mathrm{NaOH}]=0.201 \mathrm{~M}=\frac{0.201 \mathrm{~mol} \mathrm{~NaOH}}{1 \mathrm{~L} \mathrm{~NaOH}}} \\
&\text { Vol. standard }=42.0 \mathrm{~mL}=32.3 \mathrm{~mL} \mathrm{~NaOH} \times \frac{1 \mathrm{~L} \mathrm{~NaOH}}{1000 \mathrm{~mL} \mathrm{~NaOH}}=0.0323 \mathrm{~L} \mathrm{~NaOH} \\
&\text { Vol. analyte }=50.0 \mathrm{~mL}=50.0 \mathrm{~mL~HCl} \times \frac{1 \mathrm{~L} \mathrm{~HCl}}{1000 \mathrm{~mL} \mathrm{~HCl}}=0.0500 \mathrm{~L} \mathrm{~HCl}
\end{aligned}\nonumber
\end{equation}
Step 2) Calculate the moles of the standard from the volume and the molarity product.
\begin{equation}
\text { Moles of } \mathrm{~NaOH} \text { consumed }=0.0323 \mathrm{~N} \mathrm{~NaOH} \times \frac{0.201 \mathrm{~mol} \mathrm{~NaOH}}{1 \mathrm{~L} \mathrm{~NaOH}}=0.00649 \text {~mol } {~NaOH }\nonumber
\end{equation}
Step 3) write the balanced chemical equation of the reaction.
\begin{equation}
\mathrm{H}_{2} \mathrm{SO}_{4}(\mathrm{aq})+2 \mathrm{NaOH}(\mathrm{aq}) \rightarrow \mathrm{Na}_{2} \mathrm{SO}_{4}(\mathrm{aq})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l})\nonumber
\end{equation}
Step 4) write the conversion factor for the moles of standard to mole of analyte calculation from the equation.
\begin{equation}
\frac{1 \mathrm{~mol}\mathrm{~H}_{2} \mathrm{SO}_{4}}{2 \mathrm{~mol} \mathrm{~NaOH}}\nonumber
\end{equation}
Step 5) Calculate the moles of the analyte by multiplying the moles of the standard and the conversion factor.
\begin{equation}
0.00649 \text { mol } \mathrm{~NaOH} \times \frac{1 \mathrm{~mol} \mathrm{~H}_{2} \mathrm{SO}_{4}}{2 \mathrm{~mol} \mathrm{~NaOH}}=0.00325 \mathrm{~mol} \mathrm{~H}_{2} {SO}_{4}\nonumber
\end{equation}
Step 6) Calculate the molarity of the analyte by dividing the moles with the volume in liters of the analyte.
\begin{equation}
\text { Concentration of } \mathrm{~H}_{2} \mathrm{SO}_{4}=\frac{0.00325 \mathrm{~mol} \mathrm{~H}_{2} \mathrm{SO}_{4}}{0.0500 \mathrm{~L} \mathrm{~H}_{2} \mathrm{SO}_{4}}=0.0649 \mathrm{~M} {~H_{2}} SO_{4}\nonumber
\end{equation}
The same calculation in a summary form is in the following.
\[\text {Concentration of} \mathrm{~H}_{2} \mathrm{SO}_{4}=\frac{0.00325 \mathrm{~mol} \mathrm{~H}_{2} \mathrm{SO}_{4}}{0.0500 \mathrm{~L~H}_{2} \mathrm{SO}_{4}}=0.0649 \mathrm{~M} \mathrm{~H}_{2} \mathrm{SO}_{4}\nonumber\]
\[\text{Moles of} \mathrm{~H}_{2} \mathrm{SO}_{4} =\quad 32.3 \mathrm{~mL} \mathrm{~NaOH} \times \frac{1 \mathrm{~L} \mathrm{~NaOH}}{1000 \mathrm{~mL} \mathrm{~NaOH}} \times \frac{0.201 \mathrm{~mol} \mathrm{~NaOH}}{1 \mathrm{~L} \mathrm{~NaOH}} \times \frac{1 \mathrm{~mol} \mathrm{~H}_{2} \mathrm{SO}_{4}}{2 \mathrm{~mol} \mathrm{~NaOH}}=0.00326 \mathrm{~mol} \mathrm{} \mathrm{~H}_{2} \mathrm{SO}_{4}\nonumber\]
\[\text{Molarity of} \mathrm{~H}_{2} \mathrm{SO}_{4}: \quad M=\frac{n(\mathrm{~mol})}{V(L)}=\frac{0.00326 \mathrm{~mol} \mathrm{~H}_{2} \mathrm{SO}_{4}}{50.0 \mathrm{~mL} \mathrm{~H}_{2} \mathrm{SO}_{4}} \times \frac{1000 \mathrm{~mL} \mathrm{~H}_{2} \mathrm{SO}_{4}}{1 \mathrm{~L~H}_{2} \mathrm{SO}_{4}}=0.0649 \mathrm{~M} \mathrm{~H}_{2} \mathrm{SO}_{4}\nonumber\]


11: Biomolecules- Carbohydrates
The chapter begins with an explanation of the classification schemes used to simplify the study of carbohydrates. We make extensive use of Fischer projection formulas throughout the chapter. We place considerable emphasis on gaining an appreciation of the configurations of carbohydrates, particularly of the aldoses. We describe the disadvantages of representing monosaccharides by open-chain structures, and at this point, introduce you to cyclic representations—called Haworth projections—of these substances. We describe the mutarotation of glucose, explaining it in terms of the existence of anomers. We then examine some reactions of monosaccharides, including the formation of ethers and esters, the formation of glycosides, and reduction and oxidation. We discuss the structures of some common disaccharides and polysaccharides, and conclude the chapter with a brief explanation of the role played by carbohydrates in cell recognition.
-
- 11.3: Depicting Carbohydrate Stereochemistry - Fischer Projections
- The wedge and dash notations we have been using are effective for drawing three-dimensional configurations on a two-dimensional surface,, but can be troublesome when applied to compounds having many chiral centers. Emil Fischer devised a simple notation that is still widely used. Fischer Projections allow us to represent 3D molecular structures in a 2D environment without changing their properties and/or structural integrity.
-
- 11.7: Reactions of Monosaccharides
- Monosaccharides contain both alcohol and carbonyl functional groups. This allows monosaccharides to undergo many of the reactions typical for these functional groups. In particular, alcohols can be converted to esters, converted to ethers, converted to acetals, or oxidized. Carbonyls can be reacted with nucleophiles, be reduced to form alcohols, or be oxidized to form carboxylic acids.


20.30: Exercise and Memory
-
- Last updated
- Save as PDF
Learning Objectives
- To study the benefits of exercise on memory
Research conducted by
M. E. Hopkins, F. C. Davis, M. R. Van Tieghem, P. J. Whalen, and D. J. Bucci
Case study prepared by
Robert F. Houser and Georgette Baghdady
Overview
Physical exercise has many beneficial effects on physiological processes, including those that affect cognition and memory. Exercise increases brain-derived neurotrophic factor ( BDNF ), which is a protein found in the learning and memory centers of the brain where it supports nerve cell survival and the growth of new neurons and neuronal connections. A polymorphism of BDNF (a variant genotype) alters the release of BDNF during exercise. The researchers of this study sought to compare the effects of a single bout of exercise versus a \(4\)-week exercise regimen on cognition and memory and to determine if BDNF genotype influences the intensity of those effects of exercise.
Questions to Answer
How do regular exercise and/or an acute bout of exercise affect cognitive memory? Does type of BDNF genotype (Val/Val or Met carrier) mediate the effect of exercise on memory? How do we calculate a one-way ANOVA by hand and how do different post-hoc tests compare?
Design Issues
The group sample sizes are small, perhaps limiting the power to detect significant differences between the four exercise/control groups.
Descriptions of Variables
| Variable | Description |
| Group |
0W-: sedentary group
0W+: sedentary group with one bout of exercise at least 2 hours before Visit 2 4W-: regularly exercising group 4W+: regularly exercising group with a bout of exercise at least 2 hours before Visit 2 |
|
Accuracy |
The percentage of objects each group accurately identified as old or new when performing the novel object recognition task during each study visit |
| Difference score | Accuracy achieved by the subject in the novel object recognition task during Visit 2 minus accuracy during Visit 1, in percent |
| BDNF genotype | Whether a subject's BDNF genotype is Val/Val or Met carrier (Val/Met and Met/Met) |
Links
How Exercise Affects the Brain: Age and Genetics Play a Role
BDNF
References
- Hopkins, M. E., Davis, F. C., Van Tieghem, M. R., Whalen, P. J., Bucci, D. J. (2012). Differential effects of acute and regular physical exercise on cognition and affect. Neuroscience, 215, 59-68.


1.1: Module 1- Viewing 3D Models – Part 1
Module 1: Viewing 3D Models – Part 1
Learning Outcomes
When you have completed this module, you will be able to:
- Describe AutoCAD’s Ribbon and Toolbar menu systems and set one to used for this book.
- Describe a wireframe model and explain how it is viewed in 3D.
- Describe the UCS icon and explain how to configure and display it.
- Apply the VIEW, 3DORBIT, 3DFORBIT, 3DCORBIT, and UCSICON commands.
AutoCAD’s Menu Systems
AutoCAD has two menu systems that allows you to communicate with it. The two menu systems are Toolbar menu and Ribbon menu. See Figure 1-1. You can, and should, add the Pull-down menu to either the Toolbar or the Ribbon menu. You can use either menu system when completing labs exercises in this book. You can switch between them, if you wish.
WORK ALONG: Setting AutoCAD’s Interface Using Toolbar Menus
If you are using Ribbon menus, skip to WORK ALONG: Setting AutoCAD’s Interface Using Ribbon Menus
Step 1
Launch AutoCAD. AutoCAD’s blank window will display. (Figure Step 1)
Step 2
Click the New icon. This will open the Select Template dialogue box. (Figure Step 2)
Step 3
In the Select template dialogue box, click the template file: 3D Layout English to highlight it. Then click the Open button. This will open AutoCAD’s Graphic window. (Figure Step 3)
Step 4
Click the small arrow in the Workplace Switching icon on the Status bar. Click the workspace: AutoCADbook to set it as the current workspace. Make sure that the check mark is beside the workspace: AutoCADbook as shown in the figure. (Figure Step 4A and 4B)
Step 5
Check to ensure that the tab Model is enabled. It is located on the bottom left corner of the Graphic window. If it is not enabled, click it with the left mouse button. (Figure Step 5)
Step 6
Disable all features on the Status bar by clicking any that display with a blue background as shown in Figure Step 6A. All features should display with a gray background as shown in Figure Step 6B. (Figure Step 6A and 6B)
USER TIP: You can easily check to see if a feature on the Status bar is enabled or disabled by holding your graphic cursor on the icon. A pop up window will display indicating the name of the feature and whether it is currently on or off as shown in the figures below.
Step 7
Move the Graphic cursor onto any toolbar icon and right click the mouse. This will pull down a list of all available toolbar menus. The ones that are preceded with a checkmark are enabled and are currently displayed in the Graphic window. Ensure that the toolbars: Layers, Orbit, Properties, Standard, Styles, UCS, UCS II , View , and Visual Styles are the only ones enabled. Click on the name to toggle the display of the toolbar. (Figure Step 7)
Step 8
Your Graphic screen should match the figure very closely. (Figure Step 8)
Step 9
Enter the MENUBAR command. If is set to 0, set it to 1, as shown below.
Command: MENUBAR
Enter new value for MENUBAR <0>: 1
Command:
Step 10
Click File on the Pull-down menu. Click Exit to close AutoCAD. If you are asked if you want to save the drawing, click No. (Figure Step 10)
Step 11
Skip to Viewing in 3D
WORK ALONG: Setting AutoCAD’s Interface Using Ribbon Menus
If you are using Toolbar menus, go to Viewing in 3D
Step 1
Launch AutoCAD. AutoCAD’s blank window will display. (Figure Step 1)
Step 2
Click the New icon. This will open the Select Template dialogue box. (Figure Step 2)
Step 3
In the Select Template dialogue box, click the template file: 3 D Layout English to highlight it. Then click the Open button. This will open AutoCAD’s Graphic window. (Figure Step 3)
Step 4
Click the small arrow in the Workplace Switching icon on the Status bar. Click the workspace: 3D Modeling to set it as the current workspace. Ensure that the check mark is beside the workspace: 3D Modeling as shown in the figure. (Figure Step 4A and 4B)
Step 5
Enter the MENUBAR command and if is set to 0, set it to 1, as shown below. Command: MENUBAR
Enter new value for MENUBAR <0>: 1
Command:
Step 6
Check to ensure that the tab Model is enabled. If it is not enabled, click it with the left mouse button. (Figure Step 6)
Step 7
Disable all features on the Status bar by clicking any that display with a blue background as shown in Figure Step 7A. All features should display with a gray background as shown in Figure Step 7B. (Figure Step 7A and 7B)
Step 8
Click File on the Pull-down menu. Click Exit to close AutoCAD. If you are asked if you want to save the drawing, click No. (Figure Step 8)
USER TIP: You can easily check to see if a feature on the Status bar is enabled or disabled by holding yourgraphic cursor on the icon. A pop up window will display indicating the name of the feature and whether it is currently on or off as shown in the figures on the right.
Viewing in 3D
While working in 2D (two dimensions), the model was always viewed looking down from the top. The only viewing adjustment made, when looking at the model, was the distance (zoom) and the location (pan). There was no need to change the viewing angle. Working in 3D (three dimensions) is quite different. While the model can still be zoomed and panned, you can change the viewing direction and angle of the model.
Throughout the AutoCAD 2D book, the objects being drawn and modified were called the drawing objects even though they were actually model objects on the top view of a model. The reason this was done was to keep it simple and allow you to concentrate on learning to draw and modify geometry. In the AutoCAD 3D book, all objects that are drawn in model space will be referred to as model objects and all objects drawn in paper space will be called drawing objects.
UCS Icon
The User Coordinate System (UCS) icon displays the location and orientation of the current coordinate system. Its location, which is controlled by you, is at either the current coordinate system origin point or in the lower-left corner of the current viewport. When constructing a model, it is best to locate it at the origin point as taught in Modules 3 and 4. Figure 1-2 shows the two different icons as they would appear in a 3D viewing orientation.
A Wireframe Model
A model is the 3 dimensional object drawn at full size and located accurately in relation to model space origin or X0Y0Z0 in the World UCS. A wireframe model is a real-world 3D object represented by lines, circles, arcs, and/or plines located along the edge of the model. See Figure 1-3. A wireframe model is hollow in the centre. That is the reason that lines and arcs on the opposite side of the model can be seen. Think of it as the skeleton of an object. The same model when surfaced or constructed as a solid model will appear as shown in Figure 1-4.
Surfacing a wireframe model and constructing a solid model are covered in Modules 12 to 30.
Wireframe models can be used to view the model from any vantage point, generate standard views in a drawing, generate exploded and perspective views, plus can be used as a skeleton when constructing the surfaced model. A wireframe model cannot be shaded or rendered. A wireframe model can also be used to analyze spatial relationships, including the shortest distance between corners and edges and checking for interferences.
It is important to plan, organize, and construct your model using layers and colors. This will allow you to more easily visualize a complex model and differentiate between objects in various views.
A View
A view is an area displayed by the viewing angle and direction and the location of the target. The current view is the view of the model that is being displayed in the Graphic window. There are many ways of setting the current view of the model and all of them will be covered throughout the AutoCAD 3D book. In this module, using the VIEW command is taught to either select a preset view or set and name user defined views. The basic principles of the orbit commands are also covered.
AutoCAD Command: VIEW
The VIEW command is used to display the preset views or to define, name, and display user defined views of the model.
Shortcut: none
AutoCAD Commands: 3DORBIT, 3DFORBIT, 3DCORBIT
The 3DORBIT, 3DFORBIT, and 3DCORBIT commands allows the user to dynamically orbit the
model changing the viewing angle, viewing direction, and target of the view.
Shortcut: none
AutoCAD Commands: UCSICON
The UCSICON command is used to configure and control the display of the UCS icon.
Shortcut: none
WORK ALONG: Using the UCSICON, VIEW, and 3DFORBIT Commands
Step 1
Using the NEW command, start a new drawing using the template: 3D Layout English .
Step 2
Save and name the drawing: AutoCAD 3D Workalong 01-1.
Step 3
In the View pull-down, select Display – UCS Icon – Properties. Ensure that both On and Origin enabled as shown in the figure. This will open the UCS Icon dialogue box. Ensure that the settings are the same in your dialogue box as they are in the figure. (Figure Step 3A and 3B)
Step 4
Enter the UNITS command. In the Units dialogue box, set the Insertion Units to Inches. Using the INSERT command, insert the block: AutoCAD 3D Lab 01-1 at the coordinates 0,0,0. Explode the block that you just inserted. Your model should appear as shown in the figure. (Figure Step 4)
Step 5
If you are using Toolbar menus, use the View toolbar or if you are using Ribbon menus, use the View section of the Home tab to complete Steps 7 and 8. (Figure Step 5A and 5B)
Step 6
Enter the system variable UCSORTHO, as shown below, and set it to 0.
Command: UCSORTHO
Enter new value for UCSORTHO <1>: 0
Command:
Step 7
Click the Top view icon. Note how the model is now being viewed from the top as shown in the figure. Continue to change the view to display the ten predefined views. Note how the view of the model changes each time you click an icon. (Figure Step 7A, 7B, 7C, 7D, 7E, 7F, 7G, 7H, 7I, and 7J)
Step 8
Change the current view to SE Isometric as shown in the figure. (Figure Step 8)
Step 9
Enter the 3DFORBIT command using either the toolbar or the pull-down menu. It will display the arcball around the model. Zoom and pan the model to locate it in the centre of the arcball. (Figure Step 9A, 9B, and 9C)
Step 10
Move the cursor onto the small circle on the left side. Press and hold down the left mouse button and while holding it down, move the cursor slowly to the right. until your model appears similar to the figure. You do not have to move it very far. (Figure Step 10)
Step 11
Move the cursor onto the small circle on the bottom. Press and hold down the left mouse button and while holding it down, move the slowly up until your model appears as shown in the figure. (Figure Step 11)
Step 12
Move the cursor inside the arcball. Press and hold down the left mouse button and while holding it down, move it until the model appears as shown in the figure. (Figure Step 12)
Step 13
Press Esc to exit the command. Change the current view to SE Isometric. (Figure Step 13)
Step 14
Using 3DFORBIT, orbit the model to appear similar to the figure. It does not have to match exactly. (Figure Step 14)
Step 15
Enter the VIEW command using either the toolbar or ribbon. This will open the View Manager dialogue box. (Figure Step 15A and 15B)
Step 16
In the Views box, click Model Views to highlight it as shown in the figure and then click the New icon. This will open the New View dialogue box. In the View name box, enter the name: Module 1 View . Click OK to see the view name as shown in the figure. Click OK to close the dialogue box. (Figure Step 16A, 16B, and 16C)
Step 17
Change the current view to SE Isometric. (Figure Step 17)
Step 18
Save and close the drawing.
Key Principles
Key Principles in Module 1
- A wireframe model is a real-world 3D object with lines, circles, arcs, and/or plines located along the edge of the A wireframe model is a hollow object.
- A model is drawn in model space. It is a 3 dimensional object drawn at full size and located accurately in relation to model space origin X0Y0Z0 in the World UCS.
- A view is simply the way you are looking at the model. It is the model displayed by the viewing angle, viewing direction, and the location of the target.
- The SE Isometric view is the home view used to keep a mental view of the model.
- Do not over-orbit the model using the orbit It best to retain a good mental view of the model at all times during model construction.


8.9: Complex Rational Expressions
Simple And Complex Fractions
Simple Fraction
In section 8.2 we saw that a simple fraction was a fraction of the form \(\dfrac{P}{Q}\), where \(P\) and \(Q\) are polynomials and \(Q \not = 0\).
Complex Fraction
A complex fraction is a fraction in which the numerator or denominator, or both, is a fraction. The fractions
\(\dfrac{\frac{8}{15}}{\frac{2}{3}}\) and \(\dfrac{1 - \frac{1}{x}}{1 - \frac{1}{x^2}}\)
are examples of complex fractions, or more generally, complex rational expressions.
There are two methods for simplifying complex rational expressions: the combine-divide method and the LCD-multiply-divide method.
The Combine-Divide Method
- If necessary, combine the terms of the numerator together.
- If necessary, combine the terms of the denominator together.
- Divide the numerator by the denominator.
Sample Set A
Simplify each complex rational expression.
\(\dfrac{\frac{x^3}{8}}{\frac{x^5}{12}}\)
Steps 1 and 2 are not necessary so we proceed with step 3:
\(\dfrac{\frac{x^3}{8}}{\frac{x^5}{12}} = \dfrac{x^3}{8} \cdot \dfrac{12}{x^5} = \dfrac{\cancel{x^3}}{^\cancel{8}_2} \cdot \dfrac{_\cancel{12}^3}{x^{\cancel{5}2}} = \dfrac{3}{2x^2}\)
\(\dfrac{1 - \frac{1}{x}}{1 - \frac{1}{x^2}}\)
Step 1: Combine the terms of the numerator: LCD = \(x\).
\(1 - \dfrac{1}{x} = \dfrac{x}{x} - \dfrac{1}{x} = \dfrac{x-1}{x}\)
Step 2: Combine the terms of the denominator: LCD = \(x^2\).
\(1 - \dfrac{1}{x^2} = \dfrac{x^2}{x^2} - \dfrac{1}{x^2} = \dfrac{x^2 - 1}{x^2}\)
Step 3: Divide the numerator by the denominator.
\(\begin{array}{flushleft}
\dfrac{\frac{x-1}{x}}{\frac{x^2-1}{x^2}} &= \dfrac{x-1}{x} \cdot \dfrac{x^2}{x^2-1}\\
&= \dfrac{\cancel{x-1}}{\cancel{x}} \dfrac{x^{\cancel{2}}}{(x+1)\cancel{(x+1)}}\\
&= \dfrac{x}{x+1}
\end{array}\)
Thus,
\(\dfrac{1 - \frac{1}{x}}{1 - \frac{1}{x^2}} = \dfrac{x}{x+1}\)
\(\dfrac{2 - \frac{13}{m} - \frac{7}{m^2}}{2 + \frac{3}{m} + \frac{1}{m^2}}\)
Step 1: Combine the terms of the numerator: LCD = \(m^2\).
\(2-\dfrac{13}{m}-\dfrac{7}{m^{2}}=\dfrac{2 m^{2}}{m^{2}}-\dfrac{13 m}{m^{2}}-\dfrac{7}{m^{2}}=\dfrac{2 m^{2}-13 m-7}{m^{2}}\)
Step 2: Combine the terms of the denominator: LCD = \(m^2\)
\(2+\dfrac{3}{m}+\dfrac{1}{m^{2}}=\dfrac{2 m^{2}}{m^{2}}+\dfrac{3 m}{m^{2}}+\dfrac{1}{m^{2}}=\dfrac{2 m^{2}+3 m+1}{m^{2}}\)
Step 3: Divide the numerator by the denominator:
\(\begin{array}{flushleft}
\dfrac{\frac{2 m^{2}-13 m-7}{m^{2}}}{\frac{2 m^{2}+3 m-1}{m^{2}}} &=\dfrac{2 m^{2}-13 m-7}{m^{2}} \cdot \frac{m^{2}}{2 m^{2}+3 m+1} \\
&=\dfrac{\cancel{(2 m+1)}(m-7)}{\cancel{m^2}} \cdot \dfrac{\cancel{m^2}}{\cancel{(2 m+1)}(m+1)} \\
&=\dfrac{m-7}{m+1}
\end{array}\)
Thus,
\(\dfrac{2 - \frac{13}{m} - \frac{7}{m^2}}{2 + \frac{3}{m} + \frac{1}{m^2}} = \dfrac{m - 7}{m + 1}\)
Practice Set A
Use the combine-divide method to simplify each expression.
\(\dfrac{\frac{27x^2}{6}}{\frac{15x^3}{8}}\)
- Answer
-
\(\dfrac{12}{5x}\)
\(\dfrac{3 - \frac{1}{x}}{3 + \frac{1}{x}}\)
- Answer
-
\(\dfrac{3x - 1}{3x + 1}\)
\(\dfrac{1 + \frac{x}{y}}{x - \frac{y^2}{x}}\)
- Answer
-
\(\dfrac{x}{y(x-y)}\)
\(\dfrac{m - 3 + \frac{2}{m}}{m - 4 + \frac{3}{m}}\)
- Answer
-
\(\dfrac{m-2}{m-3}\)
\(\dfrac{1 + \frac{1}{x-1}}{1 - \frac{1}{x-1}}\)
- Answer
-
\(\dfrac{x}{x-2}\)
The LCD-Multiply-Divide Method
- Find the LCD of all the terms.
- Multiply the numerator and denominator by the LCD.
- Reduce if necessary.
Sample Set B
Simplify each complex fraction.
\(\dfrac{1 - \frac{4}{a^2}}{1 + \frac{2}{a}}\)
Step 1: The LCD \(=a^2\).
Step 2: Multiply both the numerator and denominator by \(a^2\).
\(\begin{array}{flushleft}
\dfrac{a^2(1 - \frac{4}{a^2})}{a^2(1 + \frac{2}{a})} &= \dfrac{a^2 \cdot 1-a^2 \cdot \frac{4}{a^2}}{a^2 \cdot 1+a^2\cdot\frac{2}{a}}\\
&= \dfrac{a^2-4}{a^2 + 2a}
\end{array}\).
Step 3: Reduce:
\(\begin{array}{flushleft}
\frac{a^{2}-4}{a^{2}+2 a} &=\frac{\cancel{(a+2)}(a-2)}{a\cancel{(a+2)}} \\
&=\frac{a-2}{a}
\end{array}\)
Thus,
\(\dfrac{1-\frac{4}{a^2}}{1 + \frac{2}{a}} = \dfrac{a-2}{a}\)
\(\dfrac{1 - \frac{5}{x} - \frac{6}{x^2}}{1 + \frac{6}{x} + \frac{5}{x^2}}\)
Step 1: The LCD is \(x^2\).
Step 2: Multiply the numerator and denominator by \(x^2\).
\(\begin{array}{flushleft}
\dfrac{x^{2}(1-\frac{5}{x}-\frac{6}{x^{2}})}{x^{2}(1+\frac{6}{x}+\frac{5}{x^{2}})} &= \dfrac{x^{2} \cdot 1-x^{\cancel{2}} \cdot \frac{5}{\cancel{x}}-\cancel{x^{2}} \cdot \frac{6}{\cancel{x^{2}}}}{x^{2} \cdot 1+x^{\cancel{2}} \cdot \frac{6}{\cancel{x}}+\cancel{x^2} \cdot \frac{5}{\cancel{x^2}}} \\
&=\dfrac{x^{2}-5 x-6}{x^{2}+6 x+5}
\end{array}\)
Step 3: Reduce:
\(\begin{array}{flushleft}
\dfrac{x^{2}-5 x-6}{x^{2}+6 x+5} &=\dfrac{(x-6)(x+1)}{(x+5)(x+1)} \\
&=\dfrac{x-6}{x+5}
\end{array}\)
Thus,
\(\dfrac{1 - \frac{5}{x} - \frac{6}{x^2}}{1 + \frac{6}{x} + \frac{5}{x^2}} = \dfrac{x-6}{x+5}\)
Practice Set B
The following problems are the same problems as the problems in Practice Set A. Simplify these expressions using the LCD-multiply-divide method. Compare the answers to the answers produced in Practice Set A.
\(\dfrac{\frac{27x^2}{6}}{\frac{15x^3}{8}}\)
- Answer
-
\(\dfrac{12}{5x}\)
\(\dfrac{3 - \frac{1}{x}}{3 + \frac{1}{x}}\)
- Answer
-
\(\dfrac{3x - 1}{3x + 1}\)
\(\dfrac{1 + \frac{x}{y}}{x - \frac{y^2}{x}}\)
- Answer
-
\(\dfrac{x}{y(x-y)}\)
\(\dfrac{m - 3 + \frac{2}{m}}{m - 4 + \frac{3}{m}}\)
- Answer
-
\(\dfrac{m-2}{m-3}\)
\(\dfrac{1 + \frac{1}{x-1}}{1 - \frac{1}{x-1}}\)
- Answer
-
\(\dfrac{x}{x-2}\)
Exercises
For the following problems, simplify each complex rational expression.
\(\dfrac{1+\frac{1}{4}}{1-\frac{1}{4}}\)
- Answer
-
\(\dfrac{5}{3}\)
\(\dfrac{1-\frac{1}{3}}{1+\frac{1}{3}}\)
\(\dfrac{1-\frac{1}{y}}{1+\frac{1}{y}}\)
- Answer
-
\(\dfrac{y-1}{y+1}\)
\(\dfrac{a+\frac{1}{x}}{a-\frac{1}{x}}\)
\(\dfrac{\frac{a}{b}+\frac{c}{b}}{\frac{a}{b}-\frac{c}{b}}\)
- Answer
-
\(\dfrac{a+c}{a-c}\)
\(\dfrac{\frac{5}{m}+\frac{4}{m}}{\frac{5}{m}-\frac{4}{m}}\)
\(\dfrac{3+\frac{1}{x}}{\frac{3 x+1}{x^{2}}}\)
- Answer
-
\(x\)
\(\dfrac{1+\frac{x}{x+y}}{1-\frac{x}{x+y}}\)
\(\dfrac{2+\frac{5}{a+1}}{2-\frac{5}{a+1}}\)
- Answer
-
\(\dfrac{2a + 7}{2a - 3}\)
\(\dfrac{1-\frac{1}{a-1}}{1+\frac{1}{a-1}}\)
\(\dfrac{4-\frac{1}{m^{2}}}{2+\frac{1}{m}}\)
- Answer
-
\(\dfrac{2m - 1}{m}\)
\(\dfrac{9-\frac{1}{x^{2}}}{3-\frac{1}{x}}\)
\(\dfrac{k-\frac{1}{k}}{\frac{k+1}{k}}\)
- Answer
-
\(k-1\)
\(\dfrac{\frac{m}{m+1}-1}{\frac{m+1}{2}}\)
\(\dfrac{\frac{2 x y}{2 x-y}-y}{\frac{2 x-y}{3}}\)
- Answer
-
\(\dfrac{3y^2}{(2x - y)^2}\)
\(\dfrac{\frac{1}{a+b}-\frac{1}{a-b}}{\frac{1}{a+b}+\frac{1}{a-b}}\)
\(\dfrac{\frac{5}{x+3}-\frac{5}{x-3}}{\frac{5}{x+3}+\frac{5}{x-3}}\)
- Answer
-
\(\dfrac{-3}{x}\)
\(\dfrac{2+\frac{1}{y+1}}{\frac{1}{y}+\frac{2}{3}}\)
\(\dfrac{\frac{1}{x^{2}}-\frac{1}{y^{2}}}{\frac{1}{x}+\frac{1}{y}}\)
- Answer
-
\(\dfrac{y-x}{xy}\)
\(\dfrac{1+\frac{5}{x}+\frac{6}{x^{2}}}{1-\frac{1}{x}-\frac{12}{x^{2}}}\)
\(\dfrac{1+\frac{1}{y}-\frac{2}{y^{2}}}{1+\frac{7}{y}+\frac{10}{y^{2}}}\)
- Answer
-
\(\dfrac{y-1}{y+5}\)
\(\dfrac{\frac{3 n}{m}-2-\frac{m}{n}}{\frac{3 n}{m}+4+\frac{m}{n}}\)
- Answer
-
\(3x−4\)
\(\dfrac{\frac{y}{x+y}-\frac{x}{x-y}}{\frac{x}{x+y}+\frac{y}{x-y}}\)
\(\dfrac{\frac{a}{a-2}-\frac{a}{a+2}}{\frac{2 a}{a-2}+\frac{a^{2}}{a+2}}\)
- Answer
-
\(\dfrac{4}{a^2 + 4}\)
\(3 - \dfrac{2}{1 - \frac{1}{m+1}}\)
\(\dfrac{x-\frac{1}{1-\frac{1}{x}}}{x+\frac{1}{1+\frac{1}{x}}}\)
- Answer
-
\(\dfrac{(x-2)(x+1)}{(x-1)(x+2)}\)
In electricity theory, when two resistors of resistance \(R_1\) and \(R_2\) ohms are connected in parallel, the total resistance \(R\) is:
\(R = \dfrac{1}{\frac{1}{R_1} + \frac{1}{R_2}}\)
Write this complex fraction as a simple fraction.
According to Einstein's theory of relativity, two velocities \(v_1\) and \(v_2\) are not added according to \(v = v_1 + v_2\), but rather by
\(v = \dfrac{v_1 + v_2}{1 + \frac{v_1 v_2}{c^2}}\)
Write this complex fraction as a simple fraction.
Einstein's formula is really only applicable for velocities near the speed of light (\(c=186,000\) miles per second). At very much lower velocities, such as 500 miles per hour, the formula \(v=v_1+v_2\) provides an extremely good approximation.
- Answer
-
\(\dfrac{c^2(V_1 + V_2)}{c^2 + V_1V_2}\)
Exercises For Review
Supply the missing word. Absolute value speaks to the question of how ____ and not “which way.”
Find the product. \((3x + 4)^2\)
- Answer
-
\(9x^2 + 24x + 16\)
Factor \(x^4 - y^4\)
Solve the equation \(\dfrac{3}{x-1} - \dfrac{5}{x+3} = 0\).
- Answer
-
\(x=7\)
One inlet pipe can fill a tank in 10 minutes. Another inlet pipe can fill the same tank in 4 minutes. How long does it take both pipes working together to fill the tank?


9.3: Solve Quadratic Equations by Completing the Square
By the end of this section, you will be able to:
- Complete the square of a binomial expression
- Solve quadratic equations of the form \(x^{2}+bx+c=0\) by completing the square
- Solve quadratic equations of the form \(ax^{2}+bx+c=0\) by completing the square
Before you get started, take this readiness quiz.
-
Expand: \((x+9)^{2}\).
If you missed this problem, review Example 5.32. -
Factor \(y^{2}-14 y+49\).
If you missed this problem, review Example 6.9. -
Factor \(5 n^{2}+40 n+80\).
If you missed this problem, review Example 6.14.
So far we have solved quadratic equations by factoring and using the Square Root Property. In this section, we will solve quadratic equations by a process called completing the square , which is important for our work on conics later.
Complete the Square of a Binomial Expression
In the last section, we were able to use the Square Root Property to solve the equation \((y-7)^{2}=12\) because the left side was a perfect square.
\(\begin{aligned}(y-7)^{2} &=12 \\ y-7 &=\pm \sqrt{12} \\ y-7 &=\pm 2 \sqrt{3} \\ y &=7 \pm 2 \sqrt{3} \end{aligned}\)
We also solved an equation in which the left side was a perfect square trinomial, but we had to rewrite it the form \((x−k)^{2}\) in order to use the Square Root Property.
\(\begin{aligned} x^{2}-10 x+25 &=18 \\(x-5)^{2} &=18 \end{aligned}\)
What happens if the variable is not part of a perfect square? Can we use algebra to make a perfect square?
Let’s look at two examples to help us recognize the patterns.
\(\begin{array}{cc}{(x+9)^{2}} & {(y-7)^{2}} \\ {(x+9 )(x+9)} & {(y-7)(y-7)} \\ {x^{2}+9 x+9 x+81} & {y^{2}-7 y-7 y+49} \\ {x^{2}+18 x+81} & {y^{2}-14 y+49}\end{array}\)
We restate the patterns here for reference.
If \(a\) and \(b\) are real numbers,
We can use this pattern to “make” a perfect square.
We will start with the expression \(x^{2}+6 x\). Since there is a plus sign between the two terms, we will use the \((a+b)^{2}\) pattern, \(a^{2}+2 a b+b^{2}=(a+b)^{2}\).
We ultimately need to find the last term of this trinomial that will make it a perfect square trinomial. To do that we will need to find \(b\). But first we start with determining \(a\). Notice that the first term of \(x^{2}+6x\) is a square, \(x^{2}\). This tells us that \(a=x\).
What number, \(b\) , when multiplied with \(2x\) gives \(6x\)? It would have to be \(3\), which is \(\frac{1}{2}(6)\). So \(b=3\).
Now to complete the perfect square trinomial, we will find the last term by squaring \(b\), which is \(3^{2}=9\).
We can now factor.
So we found that adding \(9\) to \(x^{2}+6 x\) ‘completes the square’, and we write it as \((x+3)^{2}\).
- Identify \(b\), the coefficient of \(x\).
- Find \(\left(\frac{1}{2} b\right)^{2}\), the number to complete the square.
- Add the\(\left(\frac{1}{2} b\right)^{2}\) to \(x^{2}+bx\).
- Factor the perfect square trinomial, writing it as a binomial squared.
Complete the square to make a perfect square trinomial. Then write the result as a binomial squared.
- \(x^{2}-26 x\)
- \(y^{2}-9 y\)
- \(n^{2}+\frac{1}{2} n\)
Solution :
a.
| The coefficient of \(x\) is -26. | |
|
Find \(\left(\frac{1}{2} b\right)^{2}\).
\(\left(\frac{1}{2} \cdot(-26)\right)^{2}\)
|
|
| Add \(169\) to the binomial to complete the square. |
\(x^{2}-26 x+169\) |
| Factor the perfect square trinomial, writing it as a binomial squared. |
\((x-13)^{2}\) |
b.
| The coefficient of \(y\) is \(-9\). | |
|
Find \(\left(\frac{1}{2} b\right)^{2}\).
\(\left(\frac{1}{2} \cdot(-9)\right)^{2}\)
|
|
| Add \(\frac{81}{4}\) to the binomial to complete the square. |
\(y^{2}-9 y+\frac{81}{4}\) |
| Factor the perfect square trinomial, writing it as a binomial squared. |
\(\left(y-\frac{9}{2}\right)^{2}\) |
c.
| The coefficient of \(n\) is \(\frac{1}{2}\). | |
|
Find \(\left(\frac{1}{2} b\right)^{2}\).
\(\left(\frac{1}{2} \cdot \frac{1}{2}\right)^{2}\)
|
|
| Add \(\frac{1}{16}\) to the binomial to complete the square. | \(n^{2}+\frac{1}{2} n+\frac{1}{16}\) |
| Rewrite as a binomial square. | \(\left(n+\frac{1}{4}\right)^{2}\) |
Complete the square to make a perfect square trinomial. Then write the result as a binomial squared.
- \(a^{2}-20 a\)
- \(m^{2}-5 m\)
- \(p^{2}+\frac{1}{4} p\)
- Answer
-
- \((a-10)^{2}\)
- \(\left(b-\frac{5}{2}\right)^{2}\)
- \(\left(p+\frac{1}{8}\right)^{2}\)
Complete the square to make a perfect square trinomial. Then write the result as a binomial squared.
- \(b^{2}-4 b\)
- \(n^{2}+13 n\)
- \(q^{2}-\frac{2}{3} q\)
- Answer
-
- \((b-2)^{2}\)
- \(\left(n+\frac{13}{2}\right)^{2}\)
- \(\left(q-\frac{1}{3}\right)^{2}\)
Solve Quadratic Equations of the Form \(x^{2}+bx+c=0\) by Completing the Square
In solving equations, we must always do the same thing to both sides of the equation. This is true, of course, when we solve a quadratic equation by completing the square too. When we add a term to one side of the equation to make a perfect square trinomial, we must also add the same term to the other side of the equation.
For example, if we start with the equation \(x^{2}+6x=40\), and we want to complete the square on the left, we will add 9 to both sides of the equation.
| Add \(9\) to both sides to complete the square. |
Now the equation is in the form to solve using the Square Root Property ! Completing the square is a way to transform an equation into the form we need to be able to use the Square Root Property.
Solve by completing the square: \(x^{2}+8x=48\).
Solution :
| Step 1 : Isolate the variable terms on one side and the constant terms on the other. | This equation has all the variables on the left. | \(\begin{array}{l}{\color{red}{x^{2}+b x\quad\:\:\: c }} \\ {x^{2}+8 x=48}\end{array}\) |
| Step 2 : Find \(\left(\frac{1}{2} \cdot b\right)^{2}\), the number to complete the square. Add it to both sides of the equation. |
Take half of \(8\) and square it. \(4^{2}=16\) Add \(16\) to BOTH sides of the equation. |
\(x^{2}+8 x+\frac{}{\color{red}{\left(\frac{1}{2} \cdot 8\right)^{2}}}\color{black}{=}48 \) \(x^{2}+8 x\color{red}{+16}\color{black}{=}48\color{red}{+16}\) |
| Step 3 : Factor the perfect square trinomial as a binomial square. |
\(x^{2}+8 x+16=(x+4)^{2}\) Add the terms on the right. |
\((x+4)^{2}=64\) |
| Step 4 : Use the Square Root Property. | \(x+4=\pm \sqrt{64}\) | |
| Step 5 : Simplify the radical and then solve the two resulting equations. |
\(x+4=\pm 8\) \(\begin{aligned} x+4 &=8 \quad x+4=-8 \\ x &=4 \quad\quad\:\:\: x=-12 \end{aligned}\) |
|
| Step 6 : Check the solutions. | Put each answer in the original equation to check. Substitute \(x=4\) and \(x=-12\). |
\(\begin{array}{r}{x^{2}+8 x=48} \\ {(\color{red}{4}\color{black}{)}^{2}+8(\color{red}{4}\color{black}{)} \stackrel{?}{=} 48} \\ {16+32\stackrel{?}{=}48} \\ {48=48}\end{array}\) \(\begin{array}{r}{x^{2}+8 x=48} \\ {(\color{red}{-12}\color{black}{)}^{2}+8(\color{red}{-12}\color{black}{)}\stackrel{?}{=} 48} \\ {144-96 \stackrel{?}{=} 48} \\ {48=48}\end{array}\) |
Solve by completing the square: \(x^{2}+4 x=5\).
- Answer
-
\(x=-5, x=-1\)
Solve by completing the square: \(y^{2}−10y=−9\).
- Answer
-
\(y=1, y=9\)
The steps to solve a quadratic equation by completing the square are listed here.
Solve a Quadratic Equation of the Form \(x^{2}+bx+c=0\) by Completing the Square
- Isolate the variable terms on one side and the constant terms on the other.
- Find \(\left(\frac{1}{2} \cdot b\right)^{2}\), the number needed to complete the square. Add it to both sides of the equation.
- Factor the perfect square trinomial, writing it as a binomial squared on the left and simplify by adding the terms on the right
- Use the Square Root Property.
- Simplify the radical and then solve the two resulting equations.
- Check the solutions.
When we solve an equation by completing the square, the answers will not always be integers.
Solve by completing the square: \(x^{2}+4 x=-21\).
Solution :
|
The variable terms are on the left side. Take half of \(4\) and square it. |
|
| \(\left(\frac{1}{2}(4)\right)^{2}=4\) | |
| Add \(4\) to both sides. | |
| Factor the perfect square trinomial, writing it as a binomial squared. | |
| Use the Square Root Property. | |
| Simplifying using complex numbers. | |
| Subtract \(2\) from each side. | |
| Rewrite to show two solutions. | |
| We leave the check to you. |
Solve by completing the square: \(y^{2}-10 y=-35\).
- Answer
-
\(y=5+\sqrt{15} i, y=5-\sqrt{15 i}\)
Solve by completing the square: \(z^{2}+8 z=-19\).
- Answer
-
\(z=-4+\sqrt{3} i, z=-4-\sqrt{3} i\)
In the previous example, our solutions were complex numbers. In the next example, the solutions will be irrational numbers.
Solve by completing the square: \(y^{2}-18 y=-6\).
Solution :
| The variable terms are on the left side. Take half of \(-18\) and square it. | |
| \(\left(\frac{1}{2}(-18)\right)^{2}=81\) | |
| Add \(81\) to both sides. | |
| Factor the perfect square trinomial, writing it as a binomial squared. | |
| Use the Square Root Property. | |
| Simplify the radical. | |
| Solve for \(y\). | |
|
Check. |
Another way to check this would be to use a calculator. Evaluate \(y^{2}−18y\) for both of the solutions. The answer should be \(−6\).
Solve by completing the square: \(x^{2}-16 x=-16\).
- Answer
-
\(x=8+4 \sqrt{3}, x=8-4 \sqrt{3}\)
Solve by completing the square: \(y^{2}+8 y=11\).
- Answer
-
\(y=-4+3 \sqrt{3}, y=-4-3 \sqrt{3}\)
We will start the next example by isolating the variable terms on the left side of the equation.
Solve by completing the square: \(x^{2}+10 x+4=15\).
Solution :
| Isolate the variable terms on the left side. Subtract \(4\) to get the constant terms on the right side. | |
| Take half of \(10\) and square it. | |
| \(\left(\frac{1}{2}(10)\right)^{2}=25\) | |
| Add \(25\) to both sides. | |
| Factor the perfect square trinomial, writing it as a binomial squared. | |
| Use the Square Root Property. | |
| Simplify the radical. | |
| Solve for \(x\). | |
| Rewrite to show two solutions. | |
| Solve the equations. | |
|
Check: |
Solve by completing the square: \(a^{2}+4 a+9=30\).
- Answer
-
\(a=-7, a=3\)
Solve by completing the square: \(b^{2}+8 b-4=16\).
- Answer
-
\(b=-10, b=2\)
To solve the next equation, we must first collect all the variable terms on the left side of the equation. Then we proceed as we did in the previous examples.
Solve by completing the square: \(n^{2}=3 n+11\).
Solution :
| Subtract \(3n\) to get the variable terms on the left side. | |
| Take half of \(-3\) and square it. | |
| \(\left(\frac{1}{2}(-3)\right)^{2}=\frac{9}{4}\) | |
| Add \(\frac{9}{4}\) to both sides. | |
| Factor the perfect square trinomial, writing it as a binomial squared. | |
| Add the fractions on the right side. | |
| Use the Square Root Property. | |
| Simplify the radical. | |
| Solve for \(n\). | |
| Rewrite to show two solutions. | |
|
Check: We leave the check for you! |
Solve by completing the square: \(p^{2}=5 p+9\).
- Answer
-
\(p=\frac{5}{2}+\frac{\sqrt{61}}{2}, p=\frac{5}{2}-\frac{\sqrt{61}}{2}\)
Solve by completing the square: \(q^{2}=7 q-3\).
- Answer
-
\(q=\frac{7}{2}+\frac{\sqrt{37}}{2}, q=\frac{7}{2}-\frac{\sqrt{37}}{2}\)
Notice that the left side of the next equation is in factored form. But the right side is not zero. So, we cannot use the Zero Product Property since it says “If \(a⋅b=0\), then \(a=0\) or \(b=0\).” Instead, we multiply the factors and then put the equation into standard form to solve by completing the square.
Solve by completing the square: \((x-3)(x+5)=9\).
Solution :
| We multiple the binomials on the left. | |
| Add \(15\) to isolate the constant terms on the right. | |
| Take half of \(2\) and square it. | |
| \(\left(\frac{1}{2} \cdot(2)\right)^{2}=1\) | |
| Add \(1\) to both sides. | |
| Factor the perfect square trinomial, writing it as a binomial squared. | |
| Use the Square Root Property. | |
| Solve for \(x\). | |
| Rewrite to show two solutions. | |
| Simplify. | |
|
Check: We leave the check for you! |
Solve by completing the square: \((c-2)(c+8)=11\).
- Answer
-
\(c=-9, c=3\)
Solve by completing the square: \((d-7)(d+3)=56\).
- Answer
-
\(d=11, d=-7\)
Solve Quadratic Equations of the Form \(ax^{2}+bx+c=0\) by Completing the Square
The process of completing the square works best when the coefficient of \(x^{2}\) is \(1\), so the left side of the equation is of the form \(x^{2}+bx+c\). If the \(x^{2}\) term has a coefficient other than \(1\), we take some preliminary steps to make the coefficient equal to \(1\).
Sometimes the coefficient can be factored from all three terms of the trinomial. This will be our strategy in the next example.
Solve by completing the square: \(3 x^{2}-12 x-15=0\).
Solution :
To complete the square, we need the coefficient of \(x^{2}\) to be one. If we factor out the coefficient of \(x^{2}\) as a common factor, we can continue with solving the equation by completing the square.
| Factor out the greatest common factor. | |
| Divide both sides by \(3\) to isolate the trinomial with coefficient \(1\). | |
| Simplify. | |
| Add \(5\) to get the constant terms on the right side. | |
| Take half of \(4\) and square it. | |
| \(\left(\frac{1}{2}(-4)\right)^{2}=4\) | |
| Add \(4\) to both sides. | |
| Factor the perfect square trinomial, writing it as a binomial squared. | |
| Use the Square Root Property. | |
| Solve for \(x\). | |
| Rewrite to show two solutions. | |
| Simplify. | |
|
Check: |
Solve by completing the square: \(2 m^{2}+16 m+14=0\).
- Answer
-
\(m=-7, m=-1\)
Solve by completing the square: \(4 n^{2}-24 n-56=8\).
- Answer
-
\(n=-2, n=8\)
To complete the square, the coefficient of the \(x^{2}\) must be \(1\). When the leading coefficient is not a factor of all the terms, we will divide both sides of the equation by the leading coefficient! This will give us a fraction for the second coefficient. We have already seen how to complete the square with fractions in this section.
Solve by completing the square: \(2 x^{2}-3 x=20\).
Solution :
To complete the square we need the coefficient of \(x^{2}\) to be one. We will divide both sides of the equation by the coefficient of \(x^{2}\). Then we can continue with solving the equation by completing the square.
| Divide both sides by \(2\) to get the coefficient of \(x^{2}\) to be \(1\). | |
| Simplify. | |
| Take half of \(-\frac{3}{2}\) and square it. | |
| \(\left(\frac{1}{2}\left(-\frac{3}{2}\right)\right)^{2}=\frac{9}{16}\) | |
| Add \(\frac{9}{16}\) to both sides. | |
| Factor the perfect square trinomial, writing it as a binomial squared. | |
| Add the fractions on the right side. | |
| Use the Square Root Property. | |
| Simplify the radical. | |
| Solve for \(x\). | |
| Rewrite to show two solutions. | |
| Simplify. | |
|
Check: We leave the check for you! |
Solve by completing the square: \(3 r^{2}-2 r=21\).
- Answer
-
\(r=-\frac{7}{3}, r=3\)
Solve by completing the square: \(4 t^{2}+2 t=20\).
- Answer
-
\(t=-\frac{5}{2}, t=2\)
Now that we have seen that the coefficient of \(x^{2}\) must be \(1\) for us to complete the square, we update our procedure for solving a quadratic equation by completing the square to include equations of the form \(a x^{2}+b x+c=0\).
- Divide by aa to make the coefficient of \(x^{2}\) term \(1\).
- Isolate the variable terms on one side and the constant terms on the other.
- Find \(\left(\frac{1}{2} \cdot b\right)^{2}\), the number needed to complete the square. Add it to both sides of the equation.
- Factor the perfect square trinomial, writing it as a binomial squared on the left and simplify by adding the terms on the right
- Use the Square Root Property.
- Simplify the radical and then solve the two resulting equations.
- Check the solutions.
Solve by completing the square: \(3 x^{2}+2 x=4\).
Solution :
Again, our first step will be to make the coefficient of \(x^{2}\) one. By dividing both sides of the equation by the coefficient of \(x^{2}\), we can then continue with solving the equation by completing the square.
| Divide both sides by \(3\) to make the coefficient of \(x^{2}\) equal \(1\). | |
| Simplify. | |
| Take half of \(\frac{2}{3}\) and square it. | |
| \(\left(\frac{1}{2} \cdot \frac{2}{3}\right)^{2}=\frac{1}{9}\) | |
| Add \(\frac{1}{9}\) to both sides. | |
| Factor the perfect square trinomial, writing it as a binomial squared. | |
| Use the Square Root Property. | |
| Simplify the radical. | |
| Solve for \(x\). | |
| Rewrite to show two solutions. | |
|
Check: We leave the check for you! |
Solve by completing the square: \(4 x^{2}+3 x=2\).
- Answer
-
\(x=-\frac{3}{8}+\frac{\sqrt{41}}{8}, x=-\frac{3}{8}-\frac{\sqrt{41}}{8}\)
Solve by completing the square: \(3 y^{2}-10 y=-5\).
- Answer
-
\(y=\frac{5}{3}+\frac{\sqrt{10}}{3}, y=\frac{5}{3}-\frac{\sqrt{10}}{3}\)
Access these online resources for additional instruction and practice with completing the square.
Key Concepts
-
Binomial Squares Pattern
If \(a\) and \(b\) are real numbers,
-
How to Complete a Square
- Identify \(b\), the coefficient of \(x\).
- Find \(\left(\frac{1}{2} b\right)^{2}\), the number to complete the square.
- Add the \(\left(\frac{1}{2} b\right)^{2}\) to \(x^{2}+bx\)
- Rewrite the trinomial as a binomial square
-
How to solve a quadratic equation of the form
\(a x^{2}+b x+c=0\)
by completing the square.
- Divide by \(a\) to make the coefficient of \(x^{2}\) term \(1\).
- Isolate the variable terms on one side and the constant terms on the other.
- Find \(\left(\frac{1}{2} \cdot b\right)^{2}\), the number needed to complete the square. Add it to both sides of the equation.
- Factor the perfect square trinomial, writing it as a binomial squared on the left and simplify by adding the terms on the right.
- Use the Square Root Property.
- Simplify the radical and then solve the two resulting equations.
- Check the solutions.