3.3: Determining the Structure of Earth
The previous section described the properties and composition of Earth’s interior, which begs the question: how can we know what conditions are like deep in the Earth? It’s easy to sample the crust through drilling, and mantle material often comes to the surface as magma, but the farthest we have been able to drill into the crust so far is only about 12 km; this for a planet with a radius of 6370 km! So to understand the composition and structure of the Earth’s deep interior, we need to use indirect methods such as seismology.
Seismology is the study of vibrations within the Earth. These vibrations are caused by various events, including earthquakes, extraterrestrial impacts, explosions, storm waves hitting the shore, and tidal effects. Of course, seismic techniques have been most widely applied to the detection and study of earthquakes, but there are many other applications, and arguably seismic waves provide the most important information that we have concerning Earth’s interior. Before going any deeper into Earth, however, we need to take a look at the properties of seismic waves. The types of waves that are useful for understanding Earth’s interior are called body waves , meaning that, unlike the surface waves on the ocean, they are transmitted through Earth materials.
Imagine hitting a large block of strong rock (e.g., granite) with a heavy sledgehammer. At the point where the hammer strikes it, a small part of the rock will be compressed by a fraction of a millimeter. That compression will transfer to the neighboring part of the rock, and so on through to the far side of the rock, from where it will bounce back to the top — all in a fraction of a second. This is known as a compression wave, and it can be illustrated by holding a loose spring (like a Slinky) that is attached to something (or someone) at the other end. If you give it a sharp push so the coils are compressed, the compression propagates (travels) along the length of the spring and back (Figure \(\PageIndex{1}\)). You can think of a compression wave as a “push” wave — it’s called a P-wave (although the “P” stands for “primary” because P-waves are the first to arrive at seismic stations). In a P-wave the motion of the particles is parallel to the direction of wave propagation.
When we hit a rock with a hammer, we also create a different type of body wave, one that is characterized by back-and-forth vibrations (as opposed to compressions). This is known as a shear wave ( S-wave , where the “S” stands for “secondary”), and an analogy would be what happens when you flick a length of rope with an up-and-down motion. As shown in Figure \(\PageIndex{1}\), a wave will form in the rope, which will travel to the end of the rope and back. In this case, the motion of the particles is perpendicular to the direction the wave travels.
Compression waves and shear waves travel very quickly through geological materials. As shown in Figure \(\PageIndex{2}\), typical P-wave velocities are between 0.5 km/s and 2.5 km/s in unconsolidated sediments, and between 3.0 km/s and 6.5 km/s in solid crustal rocks. Of the common rocks of the crust, velocities are greatest in basalt and granite. S-waves are slower than P-waves, with velocities between 0.1 km/s and 0.8 km/s in soft sediments, and between 1.5 km/s and 3.8 km/s in solid rocks.
Mantle rock is generally denser and stronger than crustal rock and both P- and S-waves travel faster through the mantle than they do through the crust. Moreover, seismic-wave velocities are related to how tightly compressed a rock is, and the level of compression increases dramatically with depth. Finally, seismic waves are affected by the phase state of rock. They are slowed if there is any degree of melting in the rock. If the material is completely liquid, P-waves are slowed dramatically and S-waves are stopped altogether.
Accurate seismometers have been used for earthquake studies since the late 1800s, and systematic use of seismic data to understand Earth’s interior started in the early 1900s. The rate of change of seismic waves with depth in the Earth (Figure \(\PageIndex{3}\)) has been determined over the past several decades by analyzing seismic signals from large earthquakes at seismic stations around the world. Small differences in arrival time of signals at different locations have been interpreted to show that:
- Velocities are greater in mantle rock than in the crust.
- Velocities generally increase with pressure, and therefore with depth.
- Velocities slow in the area between 100 km and 250 km depth (called the “low-velocity zone”; equivalent to the asthenosphere).
- Velocities increase dramatically at 660 km depth (because of a mineralogical transition).
- Velocities slow in the region just above the core-mantle boundary (the D” layer or “ultra-low-velocity zone”).
- S-waves do not pass through the outer part of the core.
-
P-wave velocities increase dramatically at the boundary between the liquid outer core and the solid inner core.
One of the first discoveries about Earth’s interior made through seismology was in the early 1900s when Croatian seismologist Andrija Mohorovičić (pronounced Moho-ro-vi-chich ) realized that at certain distances from an earthquake, two separate sets of seismic waves arrived at a seismic station within a few seconds of each other. He reasoned that the waves that went down into the mantle, traveled through the mantle, and then were bent upward back into the crust, reached the seismic station first because although they had farther to go, they traveled faster through mantle rock (as shown in Figure \(\PageIndex{4}\)). The boundary between the crust and the mantle is known as the Mohorovičić discontinuity (or Moho ). Its depth is between 60 km and 80 km beneath major mountain ranges, around 30 km to 50 km beneath most of the continental crust, and between 5 km and 10 km beneath the oceanic crust.
Our current understanding of the patterns of seismic wave transmission through Earth is summarized in Figure \(\PageIndex{5}\). Because of the gradual increase in density with depth, all waves are refracted toward the lower density, slower velocity material as they travel through homogenous parts of Earth, and thus tend to curve outward toward the surface. Waves are also refracted at boundaries within Earth, such as at the Moho, at the core-mantle boundary (CMB), and at the outer-core/inner-core boundary. S-waves do not travel through liquids — they are stopped at the CMB — and there is an S-wave shadow on the side of Earth opposite a seismic source. The angular distance from the seismic source to the shadow zone is 103° on either side, so the total angular distance of the shadow zone is 154°. We can use this information to infer the depth to the CMB. P-waves do travel through liquids, so they can make it through the liquid part of the core. Because of the refraction that takes place at the CMB, waves that travel through the core are bent away from the surface, and this creates a P-wave shadow zone on either side, from 103° to 150°. This information can be used to discover the differences between the inner and outer parts of the core.
Using data from many seismometers and hundreds of earthquakes, it is possible to create a two- or three-dimensional image of the seismic properties of part of the mantle. This technique is known as seismic tomography, and an example of the result is shown in Figure \(\PageIndex{6}\).
The Pacific Plate subducts beneath Tonga and appears in Figure \(\PageIndex{6}\) as a 100 km thick slab of cold (blue-colored) oceanic crust that has pushed down into the surrounding hot mantle. The cold rock is more rigid than the surrounding hot mantle rock, so it is characterized by slightly faster seismic velocities. There is volcanism in the Lau spreading center and also in the Fiji area, and the warm rock in these areas has slower seismic velocities (yellow and red colors).
Seismic waves provide us with the structure of the inner Earth, but what about it’s other properties? In terms of composition, there are several lines of evidence pointing to a core composed mostly of iron and nickel. Wave properties suggest the core is composed of an element with an atomic number around 25 (iron has an atomic number of 26). Aside from iron, all of the other elements with an atomic number close to 25 are too rare to make up the core. If the Earth was formed through the accretion of smaller bodies such as meteorites, we would expect the composition of Earth to be similar to the composition of meteorites. Meteorites are mostly iron and nickel, but in higher proportions than the Earth’s crust. This suggests that most of this heavy iron and nickel from the meteorites must have sunk to the Earth’s center as the planet was forming. However, the core is not dense enough to be pure iron and nickel; it it about 10% below the predicted density if that was the case. This is why scientists believe the core is composed of about 10% sulfur, oxygen, and hydrogen. Finally, if the Earth’s magnetic field comes from the fluid outer core, the outer core must contain iron. In terms of the temperatures, we can calculate the melting points of these materials over the range of pressures that they would experience in the inner Earth, and then infer the temperatures that would allow these elements to exist in their solid or liquid forms.
*”Physical Geology” by Steven Earle used under a CC-BY 4.0 international license. Download this book for free at http://open.bccampus.ca


4: Introduction to Fractions and Multiplication and Division of Fractions Last updated Save as PDF Page ID 48856 Denny Burzynski & Wade Ellis, Jr. College of Southern Nevada via OpenStax CNX 4.1: Factions of Whole Numbers 4.2: Proper Fractions, Improper Fractions, and Mixed Numbers Now that we know what positive fractions are, we consider three types of positive fractions: proper fractions, improper fractions, and mixed numbers. 4.3: Equivalent Fractions, Reducing Fractions to Lowest Terms, and Raising Fractions to Higher Terms 4.4: Multiplication of Fractions 4.5: Division of Fractions 4.6: Applications Involving Fractions 4.7: Summary of Key Concepts 4.8: Exercise Supplement 4.9: Proficiency Exam


2.12: Reading- Finding the Right Marketing Mix
How does an organization determine the right marketing mix? The answer depends on the organization’s goals. Think of the marketing mix as a recipe that can be adjusted—through small adjustments or dramatic changes—to support broader company goals.
Decisions about the marketing-mix variables are interrelated. Each of the marketing mix variables must be coordinated with the other elements of the marketing program.
Consider, for a moment, the simple selection of hair shampoo. Let’s think about three different brands of shampoo and call them Discount, Upscale, and Premium. The table below shows some of the elements of the marketing mix that impact decisions by target customers.
| Discount | Upscale | Premium | |
|---|---|---|---|
| Product | Cleansing product, pleasant smell, low-cost packaging | Cleansing product, pleasant smell, attractive packaging | Cleansing product, pleasant smell created by named ingredients, premium packaging |
| Promotion | Few, if any, broad communications | National commercials show famous female “customers” with clean, bouncy hair | Differentiating features and ingredients highlighted (e.g., safe for colored hair), as well as an emphasis on the science behind the formula. Recommended by stylist in the salon. |
| Place | Distributed in grocery stores and drugstores | Distributed in grocery stores and drugstores | Distributed only in licensed salons |
| Price | Lowest price on the shelf | Highest price in the grocery store (8 times the prices of discount) | 3 to 5 times the price of Upscale |
A number of credible studies suggest that there is no difference in the effectiveness of Premium or Upscale shampoo compared with Discount shampoo, but the communication, distribution, and price are substantially different. Each product appeals to a very different target market. Do you buy your shampoo in a grocery store or a salon? Your answer is likely based on the marketing mix that has most influenced you.
An effective marketing mix centers on a target customer. Each element of the mix is evaluated and adjusted to provide unique value to the target customer. In our shampoo example, if the target market is affluent women who pay for expensive salon services, then reducing the price of a premium product might actually hurt sales, particularly if it leads stylists in salons to question the quality of the ingredients. Similarly, making the packaging more appealing for a discount product could have a negative impact if it increases the price even slightly or if it causes shoppers to visually confuse it with a more expensive product.
The goal with the marketing mix is to align marketing activities with the needs of the target customer.
Contributors and Attributions
- Finding the Right Mix. Provided by : Lumen Learning. License : CC BY: Attribution
- Introducing Marketing. Authored by : John Burnett. Project : Global Textbook Project. License : CC BY: Attribution
- Vast Array of Hair Care Products. Authored by : Sea Turtle. Located at : https://www.flickr.com/photos/sea-turtle/3541657734/ . License : CC BY-NC-ND: Attribution-NonCommercial-NoDerivatives


1.3: Solving Quadratic Equations
Solve Quadratic Equations Using the Zero Product Property
We will first solve some quadratic equations by using the Zero Product Property. The Zero Product Property says that if the product of two quantities is zero, it must be that at least one of the quantities is zero. The only way to get a product equal to zero is to multiply by zero itself.
\(\text { If } a \cdot b=0, \text { then either } a=0 \text { or } b=0 \text { or both. }\)
We will now use the Zero Product Property , to solve a quadratic equation.
Solve: \((x+1)(x-4)=0\)
- Answer
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Solve: \((x-3)(x+5)=0\)
- Answer
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\(x=3, x=-5\)
Solve: \((y-6)(y+9)=0\)
- Answer
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\(y=6, y=-9\)
We usually will do a little more work than we did in this last example to solve the linear equations that result from using the Zero Product Property.
Solve: \((5 n-2)(6 n-1)=0\)
- Answer
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\((5 n-2)(6 n-1)=0\) Use the Zero Product Property to set
each factor to 0.\(5 n-2=0 \)
\(6 n-1=0\) Solve the equations. \(n=\frac{2}{5}\) \(n = \frac{1}{6}\) Check your answers.
Solve: \((3 m-2)(2 m+1)=0\)
- Answer
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\(m=\frac{2}{3}, m=-\frac{1}{2}\)
Solve: \((4 p+3)(4 p-3)=0\)
- Answer
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\(p=-\frac{3}{4}, p=\frac{3}{4}\)
Notice when we checked the solutions that each of them made just one factor equal to zero. But the product was zero for both solutions.
Solve: \(3 p(10 p+7)=0\)
- Answer
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\(3p(10p+7)=0\) Use the Zero Product Property to set
each factor to 0.3p=0 10p+7=0 Solve the equations. p=0 10p=−7 \(p=-\frac{7}{10}\) Check your answers.
Solve: \(2 u(5 u-1)=0\)
- Answer
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\(u=0, u=\frac{1}{5}\)
Solve: \(w(2 w+3)=0\)
- Answer
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\(w=0, w=-\frac{3}{2}\)
It may appear that there is only one factor in the next example. Remember, however, that \((y-8)^{2}\) means \((y-8)(y-8)\).
Solve: \((y-8)^{2}=0\)
- Answer
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\((y−8)^{2}=0\) Rewrite the left side as a product. (y−8)(y−8)=0 Use the Zero Product Property and
set each factor to 0.y−8=0 y−8=0 Solve the equations. y=8 y=8 When a solution repeats, we call it
a double root.Check your answer.
Solve: \((x+1)^{2}=0\)
- Answer
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\(x=1\)
Solve: \((v-2)^{2}=0\)
- Answer
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\(v=2\)
Solve Quadratic Equations by Factoring
Each of the equations we have solved in this section so far had one side in factored form. In order to use the Zero Product Property, the quadratic equation must be factored, with zero on one side. So we be sure to start with the quadratic equation in standard form, \(a x^{2}+b x+c=0\). Then we factor the expression on the left.
Solve: \(x^{2}+2 x-8=0\)
- Answer
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Solve: \(x^{2}-x-12=0\)
- Answer
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\(x=4, x=-3\)
Solve: \(b^{2}+9 b+14=0\)
- Answer
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\(b=-2, b=-7\)
- Write the quadratic equation in standard form, \(a x^{2}+b x+c=0\).
- Factor the quadratic expression.
- Use the Zero Product Property.
- Solve the linear equations.
- Check.
Before we factor, we must make sure the quadratic equation is in standard form.
Solve: \(2 y^{2}=13 y+45\)
- Answer
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\(2 y^{2}=13 y+45\) Write the quadratic equation in standard form. \(2 y^{2}-13 y-45=0\) Factor the quadratic expression. \((2 y+5)(y-9)=0\) Use the Zero Product Property
to set each factor to 0.\(2 y+5=0\) \(y-9=0\) Solve each equation. \(y=-\frac{5}{2}\) \(y=9\) Check your answers.
Solve: \(3 c^{2}=10 c-8\)
- Answer
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\(c=0, c=\frac{4}{3}\)
Solve: \(2 d^{2}-5 d=3\)
- Answer
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\(d=3, d=-\frac{1}{2}\)
Solve: \(5 x^{2}-13 x=7 x\)
- Answer
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\(5 x^{2}-13 x=7 x\) Write the quadratic equation in standard form. \(5 x^{2}-20 x=0\) Factor the left side of the equation. \(5 x(x-4)=0\) Use the Zero Product Property
to set each factor to 0.\(5x=0\) \(x−4=0\) Solve each equation. \(x=0\) \(x=4\) Check your answers.
Solve: \(6 a^{2}+9 a=3 a\)
- Answer
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\(a=0, a=-1\)
Solve: \(45 b^{2}-2 b=-17 b\)
- Answer
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\(b=0, b=-\frac{1}{3}\)
Solving quadratic equations by factoring will make use of all the factoring techniques you have learned in this chapter! Do you recognize the special product pattern in the next example?
Solve: \(144 q^{2}=25\)
- Answer
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\(\begin{array}{lrllrl} & 144 q^{2}&=&25 \\ \text { Write the quadratic equation in standard form. }& 144 q^{2}-25&=& 0 \\ \text { Factor. It is a difference of squares. } & (12 q-5)(12 q+5) & = & 0 \\ \text { Use the Zero Product Property to set each factor to } 0 . & 12 q-5&=&0 & 12 q+5&=&0 \\\text { Solve each equation. } & 12 q & = & 5 & 12 q&=&-5 \\ & q&=&\frac{5}{12} & q & =&-\frac{5}{12} \\ \text { Check your answers. }\end{array}\)
Solve: \(25 p^{2}=49\)
- Answer
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\(p=\frac{7}{5}, p=-\frac{7}{5}\)
Solve: \(36 x^{2}=121\)
- Answer
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\(x=\frac{11}{6}, x=-\frac{11}{6}\)
The left side in the next example is factored, but the right side is not zero. In order to use the Zero Product Property, one side of the equation must be zero. We’ll multiply the factors and then write the equation in standard form.
Solve: \((3 x-8)(x-1)=3 x\)
- Answer
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\(\begin{array}{ll} & (3 x-8)(x-1)=3 x \\ \text { Multiply the binomials. }& 3 x^{2}-11 x+8=3 x \\ \text { Write the quadratic equation in standard form. }& 3 x^{2}-14 x+8=0\\ \text { Factor the trinomial. }& (3 x-2)(x-4)=0\\\text { Use the Zero Product Property to set each factor to } 0 . & 3 x-2=0 \quad x-4=0 \\ \text { Solve each equation. } & 3 x=2 \quad x=4 \\ & x=\frac{2}{3} \\ \text { Check your answers. } & \text {The check is left to you! } \end{array}\)
Solve: \((2 m+1)(m+3)=12 m\)
- Answer
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\(m=1, m=\frac{3}{2}\)
Solve: \((k+1)(k-1)=8\)
- Answer
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\(k=3, k=-3\)
The Zero Product Property also applies to the product of three or more factors. If the product is zero, at least one of the factors must be zero. We can solve some equations of degree more than two by using the Zero Product Property, just like we solved quadratic equations.
Solve: \(9 m^{3}+100 m=60 m^{2}\)
- Answer
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\(\begin{array}{lrllrl} & 9 m^{3}+100 m&=&60 m^{2}\\ \text { Bring all the terms to one side so that the other side is zero. } & 9 m^{3}-60 m^{2}+100 m&=&0 \\ \text { Factor the greatest common factor first. } & m\left(9 m^{2}-60 m+100\right)&=&0 \\ \text { Factor the trinomial. } & m(3 m-10)(3 m-10)&=&0 \\ \text { Use the Zero Product Property to set each factor to 0. } & m&=&0 & 3 m-10&=&0 & 3 m-10&=&0 \\ \text { Solve each equation. } & m&=&0 & m&=&\frac{10}{3}& m&=&\frac{10}{3} \\ \text { Check your answers. } & \text { The check is left to you. } \end{array}\)
Solve: \(8 x^{3}=24 x^{2}-18 x\)
- Answer
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\(x=0, x=\frac{3}{2}\)
Solve: \(16 y^{2}=32 y^{3}+2 y\)
- Answer
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\(y=0, y=\frac{1}{4}\)
When we factor the quadratic equation in the next example we will get three factors. However the first factor is a constant. We know that factor cannot equal 0.
Solve: \(4 x^{2}=16 x+84\)
- Answer
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\(\begin{array}{lrllrl} & 4 x^{2}&=&16 x+84\\ \text { Write the quadratic equation in standard form. }& 4 x^{2}-16 x-84&=&0 \\ \text { Factor the greatest common factor first. }& 4\left(x^{2}-4 x-21\right)&=&0 \\ \text { Factor the trinomial. } & 4(x-7)(x+3)&=&0 \\ \text { Use the Zero Product Property to set each factor to 0. } & 4&\neq&0 & x-7&=&0 & x +3&=&0 \\ \text { Solve each equation. } & 4&\neq&0 & x&=&7& x&=&-3 \\ \text { Check your answers. } & \text { The check is left to you. } \end{array}\)
Solve: \(18 a^{2}-30=-33 a\)
- Answer
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\(a=-\frac{5}{2}, a=\frac{2}{3}\)
Solve: \(123 b=-6-60 b^{2}\)
- Answer
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\(b=2, b=\frac{1}{20}\)
Solve Quadratic Equations Using the Quadratic Formula
When we solved quadratic equations in the last section by completing the square, we took the same steps every time. By the end of the exercise set, you may have been wondering ‘isn’t there an easier way to do this?’ The answer is ‘yes’. Mathematicians look for patterns when they do things over and over in order to make their work easier. In this section we will derive and use a formula to find the solution of a quadratic equation.
We have already seen how to solve a formula for a specific variable ‘in general’, so that we would do the algebraic steps only once, and then use the new formula to find the value of the specific variable. Now we will go through the steps of completing the square using the general form of a quadratic equation to solve a quadratic equation for \(x\) .
We start with the standard form of a quadratic equation and solve it for \(x\) by completing the square.
| \(ax^2 + bx + c = 0, \quad a \ne 0\) | |
| Isolate the variable terms on one side. | \(ax^2 + bx \quad = -c\) |
| Make the coefficient of \(x^{2}\) equal to \(1\), by dividing by \(a\). | \(\dfrac{ax^2}{a} + \dfrac{b}{a}x \quad = -\dfrac{c}{a}\) |
| Simplify. | \(x^2+ \dfrac{b}{a}x \quad = -\dfrac{c}{a}\) |
| To complete the square, find \(\left(\dfrac{1}{2} \cdot \dfrac{b}{a}\right)^{2}\) and add it to both sides of the equation. | |
| \(\left(\dfrac{1}{2} \dfrac{b}{a}\right)^{2}=\dfrac{b^{2}}{4 a^{2}}\) | \(x^2 + \dfrac{b}{a}x +{\color{red}{\dfrac{b^2}{4a^2}}}{\color{black}{ = -\dfrac{c}{a}\,+\,}}{\color{red}{\dfrac{b^2}{4a^2}}}\) |
| The left side is a perfect square, factor it. | \(\left( x + \dfrac{b}{2a}\right)^2 = -\dfrac{c}{a}+\dfrac{b^2}{4a^2}\) |
| Find the common denominator of the right side and write equivalent fractions with the common denominator. | \(\left( x + \dfrac{b}{2a}\right)^2 =\dfrac{b^2}{4a^2}-\dfrac{c\cdot\color{red}{4a}}{a\cdot\color{red}{4a}}\) |
| Simplify. | \(\left( x + \dfrac{b}{2a}\right)^2 =\dfrac{b^2}{4a^2}-\dfrac{4ac}{4a^2}\) |
| Combine to one fraction. | \(\left( x + \dfrac{b}{2a}\right)^2 =\dfrac{b^2-4ac}{4a^2}\) |
| Use the square root property. | \(x + \dfrac{b}{2a}= \pm\sqrt{\dfrac{b^2-4ac}{4a^2}}\) |
| Simplify the radical. | \(x + \dfrac{b}{2a}= \pm\dfrac{\sqrt{b^2-4ac}}{2a}\) |
| Add \(-\dfrac{b}{2a}\) to both sides of the equation. | \(x = -\dfrac{b}{2a} \pm\dfrac{\sqrt{b^2-4ac}}{2a}\) |
| Combine the terms on the right side. | \(x = \dfrac{-b\pm\sqrt{b^2-4ac}}{2a}\) |
The final equation is called the "Quadratic Formula."
The solutions to a quadratic equation of the form \(a x^{2}+b x+c=0\), where \(a≠0\) are given by the formula:
\[x=\dfrac{-b \pm \sqrt{b^{2}-4 a c}}{2 a} \label{quad}\]
To use the Quadratic Formula , we substitute the values of \(a,b\), and \(c\) from the standard form into the expression on the right side of the formula. Then we simplify the expression. The result is the pair of solutions to the quadratic equation.
Notice the Quadratic Formula (Equation \ref{quad}) is an equation. Make sure you use both sides of the equation.
Solve by using the Quadratic Formula: \(2 x^{2}+9 x-5=0\).
Solution :
| Step 1 : Write the quadratic equation in standard form. Identify the \(a,b,c\) values. | This equation is in standard form. | \(\begin{aligned} \color{red}{a x^{2}+b x+c =0} \\ 2 x^{2}+9 x-5 =0 \\ a=2, b =9, c=-5 \end{aligned}\) |
| Step 2 : Write the quadratic formula. Then substitute in the values of \(a,b,c\). | Substitute in \(a=2, b=9, c=-5\) |
\(x=\dfrac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}\)
\(x=\dfrac{-9 \pm \sqrt{9^{2}-4 \cdot 2 \cdot(-5)}}{2 \cdot 2}\) |
| Step 3 : Simplify the fraction, and solve for \(x\). | \(\begin{array}{l}{x=\dfrac{-9 \pm \sqrt{81-(-40)}}{4}} \\ {x=\dfrac{-9 \pm \sqrt{121}}{4}} \\ {x=\dfrac{-9 \pm 11}{4}} \\ {x=\dfrac{-9+11}{4}}\quad x=\dfrac{-9-11}{4} \\ {x=\dfrac{2}{4} \quad \quad\:\:\: x=\dfrac{-20}{4}}\\ {x=\dfrac{1}{2} \quad\quad\:\:\: x=-5}\end{array}\) | |
| Step 4 : Check the solutions. | Put each answer in the original equation to check. Substitute \(x=\color{red}{\dfrac{1}{2}}\) and \(x=\color{red}{-5}\). |
\(\begin{aligned}2 x^{2}+9 x-5&=0 \\ 2\color{black}{\left(\color{red}{\dfrac{1}{2}}\right)}^{2}+9 \cdot \color{red}{\dfrac{1}{2}}\color{black}{-}5 &\stackrel{?}{=} 0 \\ 2\cdot\dfrac{1}{4}+0\cdot\dfrac{1}{2}-5&\stackrel{?}{=}0 \\ 2\cdot\dfrac{1}{4}+9\cdot\dfrac{1}{2}-5&\stackrel{?}{=}0 \\ \dfrac{1}{2}+\dfrac{9}{2}-5&\stackrel{?}{=}0 \\ \dfrac{10}{2}-5&\stackrel{?}{=}0 \\5-5&\stackrel{?}{=}0\\0&=0\end{aligned}\) \(\begin{array}{r}{2 x^{2}+9 x-5=0} \\ {2(\color{red}{-5}\color{black}{)}^{2}+9(\color{red}{-5}\color{black}{)}-5\stackrel{?}{=}0} \\ {2 \cdot 25-45-5\stackrel{?}{=}0} \\ {50-45-5\stackrel{?}{=}0} \\ {0=0}\end{array}\) |
Solve by using the Quadratic Formula: \(3 y^{2}-5 y+2=0\).
- Answer
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\(y=1, y=\dfrac{2}{3}\)
Solve by using the Quadratic Formula: \(4 z^{2}+2 z-6=0\).
- Answer
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\(z=1, z=-\dfrac{3}{2}\)
- Write the quadratic equation in standard form, \(a x^{2}+b x+c=0\). Identify the values of \(a,b\), and \(c\).
- Write the Quadratic Formula. Then substitute in the values of \(a,b\), and \(c\).
- Simplify.
- Check the solutions.
If you say the formula as you write it in each problem, you’ll have it memorized in no time! And remember, the Quadratic Formula is an EQUATION. Be sure you start with “\(x=\)”.
Solve by using the Quadratic Formula: \(x^{2}-6 x=-5\).
Solution :
|
\(x^{2}-6 x=-5\) |
|
| Write the equation in standard form by adding \(5\) to each side. |
\(x^{2}-6 x+5=0\) |
| This equation is now in standard form. |
\({\color{red}{\small{ax^2+bx + c} = \small{0}}}\)
|
| Identify the values of \(\color{cyan}a\), \(\color{red}b\), \(\color{limegreen}c\). | \({\color{cyan}a=1}\), \({\color{red}b=-6}\), \({\color{limegreen}c=5}\) |
| Write the Quadratic Formula. |
\(x=\dfrac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}\) |
| Then substitute in the values of \(a, b, c\). |
\(x=\dfrac{-\color{red} (-6 ) \color{black} \pm \sqrt{\color{red}(-6) \color{black}^{2}-4 \cdot \color{cyan}1 \color{black} \cdot ( \color{limegreen}5 \color{black})}}{2 \cdot \color{cyan} 1} \) |
| Simplify. |
\(x=\dfrac{6 \pm \sqrt{36-20}}{2}\) \(x=\dfrac{6 \pm \sqrt{16}}{2}\) \(x=\dfrac{6 \pm 4}{2}\) |
| Rewrite to show two solutions. |
\(x=\frac{6+4}{2}, \quad x=\frac{6-4}{2}\) |
| Simplify. |
\(x=\frac{10}{2}, \quad x=\frac{2}{2}\) |
| \(x=5, \quad x=1\) | |
|
Check: |
Solve by using the Quadratic Formula: \(a^{2}-2 a=15\).
- Answer
-
\(a=-3, a=5\)
Solve by using the Quadratic Formula: \(b^{2}+24=-10 b\).
- Answer
-
\(b=-6, b=-4\)
When we solved quadratic equations by using the Square Root Property, we sometimes got answers that had radicals. That can happen, too, when using the Quadratic Formula . If we get a radical as a solution, the final answer must have the radical in its simplified form.
Solve by using the Quadratic Formula: \(2 x^{2}+10 x+11=0\).
Solution :
| This equation is in standard form. | |
| Identify the values of \(a,b\) and \(c\). | |
| Write the Quadratic Formula. |
\(x=\dfrac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}\) |
| Then substitute in the values of \(a, b\), and \(c\). | |
| Simplify. |
\(x=\dfrac{-10 \pm \sqrt{100-88}}{4}\) |
|
\(x=\dfrac{-10 \pm \sqrt{12}}{4}\) |
|
| Simplify the radical. |
\(x=\dfrac{-10 \pm 2 \sqrt{3}}{4}\) |
| Factor out the common factor in the numerator. |
\(x=\dfrac{\color{red}{2}(-5 \pm \sqrt{3})}{4}\) |
| Remove the common factors. |
\(x=\dfrac{-5 \pm \sqrt{3}}{2}\) |
| Rewrite to show two solutions. |
\(x=\dfrac{-5+\sqrt{3}}{2}, \quad x=\dfrac{-5-\sqrt{3}}{2}\) |
|
Check: We leave the check for you! |
Solve by using the Quadratic Formula: \(3 m^{2}+12 m+7=0\).
- Answer
-
\(m=\dfrac{-6+\sqrt{15}}{3}, m=\dfrac{-6-\sqrt{15}}{3}\)
Solve by using the Quadratic Formula: \(5 n^{2}+4 n-4=0\).
- Answer
-
\(n=\dfrac{-2+2 \sqrt{6}}{5}, n=\dfrac{-2-2 \sqrt{6}}{5}\)
When we substitute \(a, b\), and \(c\) into the Quadratic Formula and the radicand is negative, the quadratic equation will have imaginary or complex solutions. We will see this in the next example.
Solve by using the Quadratic Formula: \(3 p^{2}+2 p+9=0\).
Solution :
| This equation is in standard form. | |
| Identify the values of \(a,b,c\). | |
| Write the Quadratic Formula. | |
| Then substitute in the values of \(a,b,c\). | |
| Simplify. | |
| Simplify the radical using complex numbers. | |
| Simplify the radical. | |
| Factor the common factor in the numerator. | |
| Remove the common factors. | |
| Rewrite in standard \(a+bi\) form. | |
| Write as two solutions. |
Solve by using the Quadratic Formula: \(4 a^{2}-2 a+8=0\).
- Answer
-
\(a=\dfrac{1}{4}+\dfrac{\sqrt{31}}{4} i, \quad a=\dfrac{1}{4}-\dfrac{\sqrt{31}}{4} i\)
Solve by using the Quadratic Formula: \(5 b^{2}+2 b+4=0\).
- Answer
-
\(b=-\dfrac{1}{5}+\dfrac{\sqrt{19}}{5} i, \quad b=-\dfrac{1}{5}-\dfrac{\sqrt{19}}{5} i\)
Remember, to use the Quadratic Formula, the equation must be written in standard form, \(a x^{2}+b x+c=0\). Sometimes, we will need to do some algebra to get the equation into standard form before we can use the Quadratic Formula.
Solve by using the Quadratic Formula: \(x(x+6)+4=0\).
Solution :
Our first step is to get the equation in standard form.
| Distribute to get the equation in standard form. | |
| This equation is now in standard form. | |
| Identify the values of \(a,b,c\). | |
| Write the Quadratic Formula. | |
| Then substitute in the values of \(a,b,c\). | |
| Simplify. | |
| Simplify the radical. | |
| Factor the common factor in the numerator. | |
| Remove the common factors. | |
| Write as two solutions. | |
|
Check: We leave the check for you! |
Solve by using the Quadratic Formula: \(x(x+2)−5=0\).
- Answer
-
\(x=-1+\sqrt{6}, x=-1-\sqrt{6}\)
Solve by using the Quadratic Formula: \(3y(y−2)−3=0\).
- Answer
-
\(y=1+\sqrt{2}, y=1-\sqrt{2}\)
When we solved linear equations, if an equation had too many fractions we cleared the fractions by multiplying both sides of the equation by the LCD. This gave us an equivalent equation—without fractions— to solve. We can use the same strategy with quadratic equations.
Solve by using the Quadratic Formula: \(\dfrac{1}{2} u^{2}+\dfrac{2}{3} u=\dfrac{1}{3}\).
Solution :
Our first step is to clear the fractions.
| Multiply both sides by the LCD, \(6\), to clear the fractions. | |
| Multiply. | |
| Subtract \(2\) to get the equation in standard form. | |
| Identify the values of \(a, b\), and \(c\). | |
| Write the Quadratic Formula. | |
| Then substitute in the values of \(a, b,\) and \(c\). | |
| Simplify. | |
| Simplify the radical. | |
| Factor the common factor in the numerator. | |
| Remove the common factors. | |
| Rewrite to show two solutions. | |
|
Check: We leave the check for you! |
Solve by using the Quadratic Formula: \(\dfrac{1}{4} c^{2}-\dfrac{1}{3} c=\dfrac{1}{12}\).
- Answer
-
\(c=\dfrac{2+\sqrt{7}}{3}, \quad c=\dfrac{2-\sqrt{7}}{3}\)
Solve by using the Quadratic Formula: \(\dfrac{1}{9} d^{2}-\dfrac{1}{2} d=-\dfrac{1}{3}\).
- Answer
-
\(d=\dfrac{9+\sqrt{33}}{4}, d=\dfrac{9-\sqrt{33}}{4}\)
Think about the equation \((x-3)^{2}=0\). We know from the Zero Product Property that this equation has only one solution, \(x=3\).
We will see in the next example how using the Quadratic Formula to solve an equation whose standard form is a perfect square trinomial equal to \(0\) gives just one solution. Notice that once the radicand is simplified it becomes \(0\), which leads to only one solution.
Solve by using the Quadratic Formula: \(4 x^{2}-20 x=-25\).
Solution :
| Add \(25\) to get the equation in standard form. | |
| Identify the values of \(a, b\), and \(c\). | |
| Write the quadratic formula. | |
| Then substitute in the values of \(a, b\), and \(c\). | |
| Simplify. | |
| Simplify the radical. | |
| Simplify the fraction. | |
|
Check: We leave the check for you! |
Did you recognize that \(4 x^{2}-20 x+25\) is a perfect square trinomial. It is equivalent to \((2 x-5)^{2}\)? If you solve \(4 x^{2}-20 x+25=0\) by factoring and then using the Square Root Property, do you get the same result?
Solve by using the Quadratic Formula: \(r^{2}+10 r+25=0\).
- Answer
-
\(r=-5\)
Solve by using the Quadratic Formula: \(25 t^{2}-40 t=-16\).
- Answer
-
\(t=\dfrac{4}{5}\)
Use the Discriminant to Predict the Number and Type of Solutions of a Quadratic Equation
When we solved the quadratic equations in the previous examples, sometimes we got two real solutions, one real solution, and sometimes two complex solutions. Is there a way to predict the number and type of solutions to a quadratic equation without actually solving the equation?
Yes, the expression under the radical of the Quadratic Formula makes it easy for us to determine the number and type of solutions. This expression is called the discriminant .
Discriminant
Let’s look at the discriminant of the equations in some of the examples and the number and type of solutions to those quadratic equations.
| Quadratic Equation (in standard form) | Discriminat \(b^{2}-4ac\) | Value of the Discriminant | Number and Type of Solutions |
|---|---|---|---|
| \(2 x^{2}+9 x-5=0\) | \(\begin{aligned} 9^{2}-& 4 \cdot 2(-5) \\ & 121 \end{aligned}\) | \(+\) | \(2\) real |
| \(4 x^{2}-20 x+25=0\) |
\((-20)^{2}-4 \cdot 4 \cdot 25\)
\(0\) |
\(0\) | \(1\) real |
| \(3 p^{2}+2 p+9=0\) |
\(2^{2}-4 \cdot 3 \cdot 9\)
\(-104\) |
\(-\) | \(2\) complex |
Using the Discriminant \(b^{2}-4ac\), to Determine the Number and Type of Solutions of a Quadratic Equation
For a quadratic equation of the form \(ax^{2}+bx+c=0\), \(a \neq 0\),
- If \(b^{2}-4 a c>0\), the equation has \(2\) real solutions.
- if \(b^{2}-4 a c=0\), the equation has \(1\) real solution.
- if \(b^{2}-4 a c<0\), the equation has \(2\) complex solutions.
Determine the number of solutions to each quadratic equation.
- \(3 x^{2}+7 x-9=0\)
- \(5 n^{2}+n+4=0\)
- \(9 y^{2}-6 y+1=0\)
Solution :
To determine the number of solutions of each quadratic equation, we will look at its discriminant.
a.
\(3 x^{2}+7 x-9=0\)
The equation is in standard form, identify \(a, b\), and \(c\).
\(a=3, \quad b=7, \quad c=-9\)
Write the discriminant.
\(b^{2}-4 a c\)
Substitute in the values of \(a, b\), and \(c\).
\((7)^{2}-4 \cdot 3 \cdot(-9)\)
Simplify.
\(49+108\)
\(157\)
Since the discriminant is positive, there are \(2\) real solutions to the equation.
b.
\(5 n^{2}+n+4=0\)
The equation is in standard form, identify \(a, b\), and \(c\).
\(a=5, \quad b=1, \quad c=4\)
Write the discriminant.
\(b^{2}-4 a c\)
Substitute in the values of \(a, b\), and \(c\).
\((1)^{2}-4 \cdot 5 \cdot 4\)
Simplify.
\(1-80\)
\(-79\)
Since the discriminant is negative, there are \(2\) complex solutions to the equation.
c.
\(9 y^{2}-6 y+1=0\)
The equation is in standard form, identify \(a, b\), and \(c\).
\(a=9, \quad b=-6, \quad c=1\)
Write the discriminant.
\(b^{2}-4 a c\)
Substitute in the values of \(a, b\), and \(c\).
\((-6)^{2}-4 \cdot 9 \cdot 1\)
Simplify.
\(36-36\)
\(0\)
Since the discriminant is \(0\), there is \(1\) real solution to the equation.
Determine the number and type of solutions to each quadratic equation.
- \(8 m^{2}-3 m+6=0\)
- \(5 z^{2}+6 z-2=0\)
- \(9 w^{2}+24 w+16=0\)
- Answer
-
- \(2\) complex solutions
- \(2\) real solutions
- \(1\) real solution
Determine the number and type of solutions to each quadratic equation.
- \(b^{2}+7 b-13=0\)
- \(5 a^{2}-6 a+10=0\)
- \(4 r^{2}-20 r+25=0\)
- Answer
-
- \(2\) real solutions
- \(2\) complex solutions
- \(1\) real solution
Identify the Most Appropriate Method to Use to Solve a Quadratic Equation
We summarize the four methods that we have used to solve quadratic equations below.
Methods for Solving Quadratic Equations
- Factoring
- Square Root Property
- Completing the Square
- Quadratic Formula
Given that we have four methods to use to solve a quadratic equation, how do you decide which one to use? Factoring is often the quickest method and so we try it first. If the equation is \(ax^{2}=k\) or \(a(x−h)^{2}=k\) we use the Square Root Property. For any other equation, it is probably best to use the Quadratic Formula. Remember, you can solve any quadratic equation by using the Quadratic Formula, but that is not always the easiest method.
What about the method of Completing the Square? Most people find that method cumbersome and prefer not to use it. We needed to include it in the list of methods because we completed the square in general to derive the Quadratic Formula. You will also use the process of Completing the Square in other areas of algebra.
Identify the Most Appropriate Method to Solve a Quadratic Equation
- Try Factoring first. If the quadratic factors easily, this method is very quick.
- Try the Square Root Property next. If the equation fits the form \(ax^{2}=k\) or \(a(x−h)^{2}=k\), it can easily be solved by using the Square Root Property.
- Use the Quadratic Formula . Any other quadratic equation is best solved by using the Quadratic Formula.
The next example uses this strategy to decide how to solve each quadratic equation.
Identify the most appropriate method to use to solve each quadratic equation.
- \(5 z^{2}=17\)
- \(4 x^{2}-12 x+9=0\)
- \(8 u^{2}+6 u=11\)
Solution :
a.
\(5z^{2}=17\)
Since the equation is in the \(ax^{2}=k\), the most appropriate method is to use the Square Root Property.
b.
\(4 x^{2}-12 x+9=0\)
We recognize that the left side of the equation is a perfect square trinomial, and so factoring will be the most appropriate method.
c.
\(8 u^{2}+6 u=11\)
Put the equation in standard form.
\(8 u^{2}+6 u-11=0\)
While our first thought may be to try factoring, thinking about all the possibilities for trial and error method leads us to choose the Quadratic Formula as the most appropriate method.
Identify the most appropriate method to use to solve each quadratic equation.
- \(x^{2}+6 x+8=0\)
- \((n-3)^{2}=16\)
- \(5 p^{2}-6 p=9\)
- Answer
-
- Factoring
- Square Root Property
- Quadratic Formula
Identify the most appropriate method to use to solve each quadratic equation.
- \(8 a^{2}+3 a-9=0\)
- \(4 b^{2}+4 b+1=0\)
- \(5 c^{2}=125\)
- Answer
-
- Quadratic Formula
- Factoring or Square Root Property
- Square Root Property
Access these online resources for additional instruction and practice with using the Quadratic Formula.
Key Concepts
- Zero Product Property If \(a \cdot b=0\), then either a=0 or b=0 or both. See Example .
-
Solve a quadratic equation by factoring
To solve a quadratic equation by factoring: See
Example
.
- Write the quadratic equation in standard form, \(a x^{2}+b x+c=0\).
- Factor the quadratic expression.
- Use the Zero Product Property.
- Solve the linear equations.
- Check.
-
Use a problem solving strategy to solve word problems
See
Example
.
- Read the problem. Make sure all the words and ideas are understood.
- Identify what we are looking for.
- Name what we are looking for. Choose a variable to represent that quantity.
- Translate into an equation. It may be helpful to restate the problem in one sentence with all the important information. Then, translate the English sentence into an algebra equation.
- Solve the equation using good algebra techniques.
- Check the answer in the problem and make sure it makes sense.
- Answer the question with a complete sentence.
Key Concepts
-
Quadratic Formula
-
The solutions to a quadratic equation of the form \(a x^{2}+b x+c=0, a \neq 0\) are given by the formula:
\(x=\dfrac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}\)
-
The solutions to a quadratic equation of the form \(a x^{2}+b x+c=0, a \neq 0\) are given by the formula:
-
How to solve a quadratic equation using the Quadratic Formula.
- Write the quadratic equation in standard form, \(a x^{2}+b x+c=0\). Identify the values of \(a, b, c\).
- Write the Quadratic Formula. Then substitute in the values of \(a, b, c\).
- Simplify.
- Check the solutions.
-
Using the Discriminant, \(b^{2}-4 a c\), to Determine the Number and Type of Solutions of a Quadratic Equation
-
For a quadratic equation of the form \(a x^{2}+b x+c=0, a \neq 0\),
- If \(b^{2}-4 a c>0\), the equation has \(2\) real solutions.
- If \(b^{2}-4 a c=0\), the equation has \(1\) real solution.
- If \(b^{2}-4 a c<0\), the equation has \(2\) complex solutions.
-
For a quadratic equation of the form \(a x^{2}+b x+c=0, a \neq 0\),
-
Methods to Solve Quadratic Equations:
- Factoring
- Square Root Property
- Completing the Square
- Quadratic Formula
-
How to identify the most appropriate method to solve a quadratic equation.
- Try Factoring first. If the quadratic factors easily, this method is very quick.
- Try the Square Root Property next. If the equation fits the form \(a x^{2}=k\) or \(a(x-h)^{2}=k\), it can easily be solved by using the Square Root Property.
- Use the Quadratic Formula. Any other quadratic equation is best solved by using the Quadratic Formula.
Glossary
- quadratic equations
- are equations in which the variable is squared.
- Zero Product Property
- The Zero Product Property states that, if the product of two quantities is zero, at least one of the quantities is zero.
Glossary
- discriminant
- In the Quadratic Formula, \(x=\dfrac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}\), the quantity \(b^{2}-4 a c\) is called the discriminant.


11.10: Chapter 9 Review Exercises
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Chapter Review Exercises
Solve Quadratic Equations Using the Square Root Property
Exercise \(\PageIndex{1}\) Solve Quadratic Equations of the Form \(ax^{2}=k\) Using the Square Root Property
In the following exercises, solve using the Square Root Property.
- \(y^{2}=144\)
- \(n^{2}-80=0\)
- \(4 a^{2}=100\)
- \(2 b^{2}=72\)
- \(r^{2}+32=0\)
- \(t^{2}+18=0\)
- \(\frac{2}{3} w^{2}-20=30\)
- \(5 c^{2}+3=19\)
- Answer
-
1. \(y=\pm 12\)
3. \(a=\pm 5\)
5. \(r=\pm 4 \sqrt{2} i\)
7. \(w=\pm 5 \sqrt{3}\)
Exercise \(\PageIndex{2}\) Solve Quadratic Equations of the Form \(a(x-h)^{2}=k\) Using the Square Root Property
In the following exercises, solve using the Square Root Property.
- \((p-5)^{2}+3=19\)
- \((u+1)^{2}=45\)
- \(\left(x-\frac{1}{4}\right)^{2}=\frac{3}{16}\)
- \(\left(y-\frac{2}{3}\right)^{2}=\frac{2}{9}\)
- \((n-4)^{2}-50=150\)
- \((4 c-1)^{2}=-18\)
- \(n^{2}+10 n+25=12\)
- \(64 a^{2}+48 a+9=81\)
- Answer
-
1. \(p=-1,9\)
3. \(x=\frac{1}{4} \pm \frac{\sqrt{3}}{4}\)
5. \(n=4 \pm 10 \sqrt{2}\)
7. \(n=-5 \pm 2 \sqrt{3}\)
Solve Quadratic Equations by Completing the Square
Exercise \(\PageIndex{3}\) Solve Quadratic Equations Using Completing the Square
In the following exercises, complete the square to make a perfect square trinomial. Then write the result as a binomial squared.
- \(x^{2}+22 x\)
- \(m^{2}-8 m\)
- \(a^{2}-3 a\)
- \(b^{2}+13 b\)
- Answer
-
1. \((x+11)^{2}\)
3. \(\left(a-\frac{3}{2}\right)^{2}\)
Exercise \(\PageIndex{4}\) Solve Quadratic Equations Using Completing the Square
In the following exercises, solve by completing the square.
- \(d^{2}+14 d=-13\)
- \(y^{2}-6 y=36\)
- \(m^{2}+6 m=-109\)
- \(t^{2}-12 t=-40\)
- \(v^{2}-14 v=-31\)
- \(w^{2}-20 w=100\)
- \(m^{2}+10 m-4=-13\)
- \(n^{2}-6 n+11=34\)
- \(a^{2}=3 a+8\)
- \(b^{2}=11 b-5\)
- \((u+8)(u+4)=14\)
- \((z-10)(z+2)=28\)
- Answer
-
1. \(d=-13,-1\)
3. \(m=-3 \pm 10 i\)
5. \(v=7 \pm 3 \sqrt{2}\)
7. \(m=-9,-1\)
9. \(a=\frac{3}{2} \pm \frac{\sqrt{41}}{2}\)
11. \(u=-6 \pm 2 \sqrt{2}\)
Solve Quadratic Equations of the Form \(ax^{2}+bx+c=0\) by Completing the Square
Exercise \(\PageIndex{5}\) Solve Quadratic Equations of the Form \(ax^{2}+bx+c=0\) by Completing the Square
In the following exercises, solve by completing the square.
- \(3 p^{2}-18 p+15=15\)
- \(5 q^{2}+70 q+20=0\)
- \(4 y^{2}-6 y=4\)
- \(2 x^{2}+2 x=4\)
- \(3 c^{2}+2 c=9\)
- \(4 d^{2}-2 d=8\)
- \(2 x^{2}+6 x=-5\)
- \(2 x^{2}+4 x=-5\)
- Answer
-
1. \(p=0,6\)
3. \(y=-\frac{1}{2}, 2\)
5. \(c=-\frac{1}{3} \pm \frac{2 \sqrt{7}}{3}\)
7. \(x=\frac{3}{2} \pm \frac{1}{2} i\)
Exercise \(\PageIndex{6}\) Solve Quadratic Equations Using the Quadratic Formula
In the following exercises, solve by using the Quadratic Formula.
- \(4 x^{2}-5 x+1=0\)
- \(7 y^{2}+4 y-3=0\)
- \(r^{2}-r-42=0\)
- \(t^{2}+13 t+22=0\)
- \(4 v^{2}+v-5=0\)
- \(2 w^{2}+9 w+2=0\)
- \(3 m^{2}+8 m+2=0\)
- \(5 n^{2}+2 n-1=0\)
- \(6 a^{2}-5 a+2=0\)
- \(4 b^{2}-b+8=0\)
- \(u(u-10)+3=0\)
- \(5 z(z-2)=3\)
- \(\frac{1}{8} p^{2}-\frac{1}{5} p=-\frac{1}{20}\)
- \(\frac{2}{5} q^{2}+\frac{3}{10} q=\frac{1}{10}\)
- \(4 c^{2}+4 c+1=0\)
- \(9 d^{2}-12 d=-4\)
- Answer
-
1. \(x=\frac{1}{4}, 1\)
3. \(r=-6,7\)
5. \(v=\frac{-1 \pm \sqrt{21}}{8}\)
7. \(m=\frac{-4 \pm \sqrt{10}}{3}\)
9. \(a=\frac{5}{12} \pm \frac{\sqrt{23}}{12} i\)
11. \(u=5 \pm \sqrt{21}\)
13. \(p=\frac{4 \pm \sqrt{5}}{5}\)
15. \(c=-\frac{1}{2}\)
Exercise \(\PageIndex{7}\) Use the Discriminant to Predict the Number of Solutions of a Quadratic Equation
In the following exercises, determine the number of solutions for each quadratic equation.
-
- \(9 x^{2}-6 x+1=0\)
- \(3 y^{2}-8 y+1=0\)
- \(7 m^{2}+12 m+4=0\)
- \(5 n^{2}-n+1=0\)
-
- \(5 x^{2}-7 x-8=0\)
- \(7 x^{2}-10 x+5=0\)
- \(25 x^{2}-90 x+81=0\)
- \(15 x^{2}-8 x+4=0\)
- Answer
-
1.
- \(1\)
- \(2\)
- \(2\)
- \(2\)
Exercise \(\PageIndex{8}\) Identify the Most Appropriate Method to Use to Solve a Quadratic Equation
In the following exercises, identify the most appropriate method (Factoring, Square Root, or Quadratic Formula) to use to solve each quadratic equation. Do not solve.
-
- \(16 r^{2}-8 r+1=0\)
- \(5 t^{2}-8 t+3=9\)
- \(3(c+2)^{2}=15\)
-
- \(4 d^{2}+10 d-5=21\)
- \(25 x^{2}-60 x+36=0\)
- \(6(5 v-7)^{2}=150\)
- Answer
-
1.
- Factor
- Quadratic Formula
- Square Root
Solve Equations in Quadratic Form
Exercise \(\PageIndex{9}\) Solve Equations in Quadratic Form
In the following exercises, solve.
- \(x^{4}-14 x^{2}+24=0\)
- \(x^{4}+4 x^{2}-32=0\)
- \(4 x^{4}-5 x^{2}+1=0\)
- \((2 y+3)^{2}+3(2 y+3)-28=0\)
- \(x+3 \sqrt{x}-28=0\)
- \(6 x+5 \sqrt{x}-6=0\)
- \(x^{\frac{2}{3}}-10 x^{\frac{1}{3}}+24=0\)
- \(x+7 x^{\frac{1}{2}}+6=0\)
- \(8 x^{-2}-2 x^{-1}-3=0\)
- Answer
-
1. \(x=\pm \sqrt{2}, x=\pm 2 \sqrt{3}\)
3. \(x=\pm 1, x=\pm \frac{1}{2}\)
5. \(x=16\)
7. \(x=64, x=216\)
9. \(x=-2, x=\frac{4}{3}\)
Solve Applications of Quadratic Equations
Exercise \(\PageIndex{10}\) Solve Applications Modeled by Quadratic Equations
In the following exercises, solve by using the method of factoring, the square root principle, or the Quadratic Formula. Round your answers to the nearest tenth, if needed.
- Find two consecutive odd numbers whose product is \(323\).
- Find two consecutive even numbers whose product is \(624\).
- A triangular banner has an area of \(351\) square centimeters. The length of the base is two centimeters longer than four times the height. Find the height and length of the base.
- Julius built a triangular display case for his coin collection. The height of the display case is six inches less than twice the width of the base. The area of the of the back of the case is \(70\) square inches. Find the height and width of the case.
- A tile mosaic in the shape of a right triangle is used as the corner of a rectangular pathway. The hypotenuse of the mosaic is \(5\) feet. One side of the mosaic is twice as long as the other side. What are the lengths of the sides? Round to the nearest tenth.
Figure 9.E.1
6. A rectangular piece of plywood has a diagonal which measures two feet more than the width. The length of the plywood is twice the width. What is the length of the plywood’s diagonal? Round to the nearest tenth.
7. The front walk from the street to Pam’s house has an area of \(250\) square feet. Its length is two less than four times its width. Find the length and width of the sidewalk. Round to the nearest tenth.
8. For Sophia’s graduation party, several tables of the same width will be arranged end to end to give serving table with a total area of \(75\) square feet. The total length of the tables will be two more than three times the width. Find the length and width of the serving table so Sophia can purchase the correct size tablecloth . Round answer to the nearest tenth.
9. A ball is thrown vertically in the air with a velocity of \(160\) ft/sec. Use the formula \(h=-16 t^{2}+v_{0} t\) to determine when the ball will be \(384\) feet from the ground. Round to the nearest tenth.
10. The couple took a small airplane for a quick flight up to the wine country for a romantic dinner and then returned home. The plane flew a total of \(5\) hours and each way the trip was \(360\) miles. If the plane was flying at \(150\) mph, what was the speed of the wind that affected the plane?
11.  The trip was \(4\) miles each way and the current was difficult. If Roy kayaked at a speed of \(5\) mph, what was the speed of the current?
12. Two handymen can do a home repair in \(2\) hours if they work together. One of the men takes \(3\) hours more than the other man to finish the job by himself. How long does it take for each handyman to do the home repair individually?
- Answer
-
2. Two consecutive even numbers whose product is \(624\) are \(24\) and \(26\), and \(−24\) and \(−26\).
4. The height is \(14\) inches and the width is \(10\) inches.
6. The length of the diagonal is \(3.6\) feet.
8. The width of the serving table is \(4.7\) feet and the length is \(16.1\) feet.
10. The speed of the wind was \(30\) mph.
12. One man takes \(3\) hours and the other man \(6\) hours to finish the repair alone.
Graph Quadratic Functions Using Properties
Exercise \(\PageIndex{11}\) Recognize the Graph of a Quadratic Function
In the following exercises, graph by plotting point.
- Graph \(y=x^{2}-2\)
- Graph \(y=-x^{2}+3\)
- Answer
-
2.
Exercise \(\PageIndex{12}\) Recognize the Graph of a Quadratic Function
In the following exercises, determine if the following parabolas open up or down.
-
- \(y=-3 x^{2}+3 x-1\)
- \(y=5 x^{2}+6 x+3\)
-
- \(y=x^{2}+8 x-1\)
- \(y=-4 x^{2}-7 x+1\)
- Answer
-
2.
- Up
- Down
Exercise \(\PageIndex{13}\) Find the Axis of Symmetry and Vertex of a Parabola
In the following exercises, find
- The equation of the axis of symmetry
-
The vertex
- \(y=-x^{2}+6 x+8\)
- \(y=2 x^{2}-8 x+1\)
- Answer
-
2. \(x=2\) ; \((2,-7)\)
Exercise \(\PageIndex{14}\) Find the Intercepts of a Parabola
In the following exercises, find the \(x\)- and \(y\)-intercepts.
- \(y=x^{2}-4x+5\)
- \(y=x^{2}-8x+15\)
- \(y=x^{2}-4x+10\)
- \(y=-5x^{2}-30x-46\)
- \(y=16x^{2}-8x+1\)
- \(y=x^{2}+16x+64\)
- Answer
-
2. \(\begin{array}{l}{y :(0,15)} \\ {x :(3,0),(5,0)}\end{array}\)
4. \(\begin{array}{l}{y :(0,-46)} \\ {x : \text { none }}\end{array}\)
6. \(\begin{array}{l}{y :(0,-64)} \\ {x :(-8,0)}\end{array}\)
Graph Quadratic Functions Using Properties
Exercise \(\PageIndex{15}\) Graph Quadratic Functions Using Properties
In the following exercises, graph by using its properties.
- \(y=x^{2}+8 x+15\)
- \(y=x^{2}-2 x-3\)
- \(y=-x^{2}+8 x-16\)
- \(y=4 x^{2}-4 x+1\)
- \(y=x^{2}+6 x+13\)
- \(y=-2 x^{2}-8 x-12\)
- Answer
-
2.
4.
6.
Exercise \(\PageIndex{16}\) Solve Maximum and Minimum Applications
In the following exercises, find the minimum or maximum value.
- \(y=7 x^{2}+14 x+6\)
- \(y=-3 x^{2}+12 x-10\)
- Answer
-
2. The maximum value is \(2\) when \(x=2\).
Exercise \(\PageIndex{17}\) Solve Maximum and Minimum Applications
In the following exercises, solve. Rounding answers to the nearest tenth.
- A ball is thrown upward from the ground with an initial velocity of \(112\) ft/sec. Use the quadratic equation \(h=-16 t^{2}+112 t\) to find how long it will take the ball to reach maximum height, and then find the maximum height.
- A daycare facility is enclosing a rectangular area along the side of their building for the children to play outdoors. They need to maximize the area using \(180\) feet of fencing on three sides of the yard. The quadratic equation \(A=-2 x^{2}+180 x\) gives the area, \(A\), of the yard for the length, \(x\), of the building that will border the yard. Find the length of the building that should border the yard to maximize the area, and then find the maximum area.
- Answer
-
2. The length adjacent to the building is \(90\) feet giving a maximum area of \(4,050\) square feet.
Graph Quadratic Functions Using Transformations
Exercise \(\PageIndex{18}\) Graph Quadratic Functions of the Form \(f(x)=x^{2}+k\)
In the following exercises, graph each function using a vertical shift.
- \(g(x)=x^{2}+4\)
- \(h(x)=x^{2}-3\)
- Answer
-
2.
Exercise \(\PageIndex{19}\) Graph Quadratic Functions of the Form \(f(x)=x^{2}+k\)
In the following exercises, graph each function using a horizontal shift.
- \(f(x)=(x+1)^{2}\)
- \(g(x)=(x-3)^{2}\)
- Answer
-
2.
Exercise \(\PageIndex{20}\) Graph Quadratic Functions of the Form \(f(x)=x^{2}+k\)
In the following exercises, graph each function using transformations.
- \(f(x)=(x+2)^{2}+3\)
- \(f(x)=(x+3)^{2}-2\)
- \(f(x)=(x-1)^{2}+4\)
- \(f(x)=(x-4)^{2}-3\)
- Answer
-
2.
4.
Exercise \(\PageIndex{21}\) Graph Quadratic Functions of the Form \(f(x)=ax^{2}\)
In the following exercises, graph each function.
- \(f(x)=2x^{2}\)
- \(f(x)=-x^{2}\)
- \(f(x)=\frac{1}{2} x^{2}\)
- Answer
-
2.
Exercise \(\PageIndex{22}\) Graph Quadratic Functions Using Transformations
In the following exercises, rewrite each function in the \(f(x)=a(x-h)^{2}+k\) form by completing the square.
- \(f(x)=2 x^{2}-4 x-4\)
- \(f(x)=3 x^{2}+12 x+8\)
- Answer
-
1. \(f(x)=2(x-1)^{2}-6\)
Exercise \(\PageIndex{23}\) Graph Quadratic Functions Using Transformations
In the following exercises,
- Rewrite each function in \(f(x)=a(x−h)^{2}+k\) form
-
Graph it by using transformations
- \(f(x)=3 x^{2}-6 x-1\)
- \(f(x)=-2 x^{2}-12 x-5\)
- \(f(x)=2 x^{2}+4 x+6\)
- \(f(x)=3 x^{2}-12 x+7\)
- Answer
-
1.
- \(f(x)=3(x-1)^{2}-4\)
-
Figure 9.E.13
3.
- \(f(x)=2(x+1)^{2}+4\)
-
Figure 9.E.14
Exercise \(\PageIndex{24}\) Graph Quadratic Functions Using Transformations
In the following exercises,
- Rewrite each function in \(f(x)=a(x−h)^{2}+k\) form
-
Graph it using properties
- \(f(x)=-3 x^{2}-12 x-5\)
- \(f(x)=2 x^{2}-12 x+7\)
- Answer
-
1.
- \(f(x)=-3(x+2)^{2}+7\)
-
Figure 9.E.15
Exercise \(\PageIndex{25}\) Find a Quadratic Function From its Graph
In the following exercises, write the quadratic function in \(f(x)=a(x−h)^{2}+k\) form.
-
Figure 9.E.16 -
Figure 9.E.17
- Answer
-
1. \(f(x)=(x+1)^{2}-5\)
Solve Quadratic Inequalities
Exercise \(\PageIndex{26}\) Solve Quadratic Inequalities Graphically
In the following exercises, solve graphically and write the solution in interval notation.
- \(x^{2}-x-6>0\)
- \(x^{2}+4 x+3 \leq 0\)
- \(-x^{2}-x+2 \geq 0\)
- \(-x^{2}+2 x+3<0\)
- Answer
-
1.
-
Figure 9.E.18 - \((-\infty,-2) \cup(3, \infty)\)
3.
-
Figure 9.E.19 - \([-2,1]\)
-
Exercise \(\PageIndex{27}\) Solve Quadratic Inequalities Graphically
In the following exercises, solve each inequality algebraically and write any solution in interval notation.
- \(x^{2}-6 x+8<0\)
- \(x^{2}+x>12\)
- \(x^{2}-6 x+4 \leq 0\)
- \(2 x^{2}+7 x-4>0\)
- \(-x^{2}+x-6>0\)
- \(x^{2}-2 x+4 \geq 0\)
- Answer
-
1. \((2,4)\)
3. \([3-\sqrt{5}, 3+\sqrt{5}]\)
5. no solution
Practice Test
Exercise \(\PageIndex{28}\)
- Use the Square Root Property to solve the quadratic equation \(3(w+5)^{2}=27\).
- Use Completing the Square to solve the quadratic equation \(a^{2}-8 a+7=23\).
- Use the Quadratic Formula to solve the quadratic equation \(2 m^{2}-5 m+3=0\).
- Answer
-
1. \(w=-2, w=-8\)
3. \(m=1, m=\frac{3}{2}\)
Exercise \(\PageIndex{29}\)
Solve the following quadratic equations. Use any method.
- \(2 x(3 x-2)-1=0\)
- \(\frac{9}{4} y^{2}-3 y+1=0\)
- Answer
-
2. \(y=\frac{2}{3}\)
Exercise \(\PageIndex{30}\)
Use the discriminant to determine the number and type of solutions of each quadratic equation.
- \(6 p^{2}-13 p+7=0\)
- \(3 q^{2}-10 q+12=0\)
- Answer
-
2. \(2\) complex
Exercise \(\PageIndex{31}\)
Solve each equation.
- \(4 x^{4}-17 x^{2}+4=0\)
- \(y^{\frac{2}{3}}+2 y^{\frac{1}{3}}-3=0\)
- Answer
-
2. \(y=1, y=-27\)
Exercise \(\PageIndex{32}\)
For each parabola, find
- Which direction it opens
- The equation of the axis of symmetry
- The vertex
- The \(x\) - and \(y\)-intercepts
-
The maximum or minimum value
- \(y=3 x^{2}+6 x+8\)
- \(y=-x^{2}-8 x+16\)
- Answer
-
2.
- down
- \(x=-4\)
- \((-4,0)\)
- \(y: (0,16); x: (-4,0)\)
- minimum value of \(-4\) when \(x=0\)
Exercise \(\PageIndex{33}\)
Graph each quadratic function using intercepts, the vertex, and the equation of the axis of symmetry.
- \(f(x)=x^{2}+6 x+9\)
- \(f(x)=-2 x^{2}+8 x+4\)
- Answer
-
2.
Exercise \(\PageIndex{34}\)
In the following exercises, graph each function using transformations.
- \(f(x)=(x+3)^{2}+2\)
- \(f(x)=x^{2}-4 x-1\)
- Answer
-
2.
Figure 9.E.21
Exercise \(\PageIndex{35}\)
In the following exercises, solve each inequality algebraically and write any solution in interval notation.
- \(x^{2}-6 x-8 \leq 0\)
- \(2 x^{2}+x-10>0\)
- Answer
-
2. \(\left(-\infty,-\frac{5}{2}\right) \cup(2, \infty)\)
Exercise \(\PageIndex{36}\)
Model the situation with a quadratic equation and solve by any method.
- Find two consecutive even numbers whose product is \(360\).
- The length of a diagonal of a rectangle is three more than the width. The length of the rectangle is three times the width. Find the length of the diagonal. (Round to the nearest tenth.)
- Answer
-
2. A water balloon is launched upward at the rate of \(86\) ft/sec. Using the formula \(h=-16 t^{2}+86 t\) find how long it will take the balloon to reach the maximum height, and then find the maximum height. Round to the nearest tenth.


4.12: ANS Medication Classes and Nursing Considerations
-
- Last updated
- Save as PDF
- Ernstmeyer & Christman (Eds.)
- Chippewa Valley Technical College via OpenRN
Classes of medication, categorized according to neuroreceptor, are further discussed in more detail below. Table 4.2 [1] contrasts agonist and antagonist medications for each ANS neuroreceptor.
Receptor |
Stimulation (Agonist) |
Inhibition (Antagonist) |
|---|---|---|
| Nicotine | Nicotine is a muscle relaxant with CNS effects. Nicotine patch is used for nicotine addiction by slowly reducing dose and avoiding withdrawal effects | Not clinically applicable |
| Muscarinic | Pilocarpine causes muscle contraction; assists with glaucoma by contracting ciliary muscle and draining fluid | Atropine in small doses inhibits secretions; in moderate doses increases heart rate; in large doses decreases gastrointestinal motility |
| Alpha-1 | Pseudoephedrine and Phenylephrine cause vasoconstriction, decreased swelling of mucus membranes, and decreased secretions | Tamsulosin relaxes smooth muscle in bladder/prostate to improve urine flow and also decreases blood pressure due to vasodilation |
| Alpha-2 | Clonidine decreases CNS outflow to treat ADHD and also reduces blood pressure and heart rate | Limited clinical use |
| Beta-1 | Dobutamine increases heart rate, force of heart contraction, and speed of conduction between SA to AV nodes | Selective B blocker: Metoprolol works on Beta-1 receptors to decrease blood pressure and heart rate |
| Beta-2 | Albuterol used for bronchodilation | Nonselective B blocker: Propranolol works on Beta-2 and Beta-1 receptors; decreases blood pressure but can also cause bronchoconstriction |
| Catecholamines stimulate multiple adrenergic receptors |
Epinephrine
and
Norepinephrine
: stimulate alpha-
and
beta-
receptors on target organs, causing increased heart rate and vasoconstriction for improved blood flow to essential organs
Dopamine has dose-dependent effects that target arteries in the kidneys, heart, and brain |
Not clinically applicable |
Supplementary Videos : See the supplementary videos below related to sympathetic and parasympathetic nervous system medications.
- This work is a derivative of Daily Med by U.S. National Library of Medicine in the public domain ↵
- Forciea, B. (2018, January 12). Sympathetic nervous system drugs. [Video]. YouTube. All rights reserved. Video used with permission. https://youtu.be/-e_s-jTPtm4 ↵
- Forciea, B. (2018, February 2). Parasympathetic nervous system drugs. [Video]. YouTube. All rights reserved. Video used with permission. https://youtu.be/ZSRk_NkbBPg ↵


11.1: Systems of Linear Equations - Two Variables
- Solve systems of equations by graphing.
- Solve systems of equations by substitution.
- Solve systems of equations by addition.
- Identify inconsistent systems of equations containing two variables.
- Express the solution of a system of dependent equations containing two variables.
A skateboard manufacturer introduces a new line of boards. The manufacturer tracks its costs, which is the amount it spends to produce the boards, and its revenue, which is the amount it earns through sales of its boards. How can the company determine if it is making a profit with its new line? How many skateboards must be produced and sold before a profit is possible? In this section, we will consider linear equations with two variables to answer these and similar questions.
Introduction to Systems of Equations
In order to investigate situations such as that of the skateboard manufacturer, we need to recognize that we are dealing with more than one variable and likely more than one equation. A system of linear equations consists of two or more linear equations made up of two or more variables such that all equations in the system are considered simultaneously. To find the unique solution to a system of linear equations, we must find a numerical value for each variable in the system that will satisfy all equations in the system at the same time. Some linear systems may not have a solution and others may have an infinite number of solutions. In order for a linear system to have a unique solution, there must be at least as many equations as there are variables. Even so, this does not guarantee a unique solution.
In this section, we will look at systems of linear equations in two variables, which consist of two equations that contain two different variables. For example, consider the following system of linear equations in two variables.
\[\begin{align*} 2x+y &= 15 \\ 3x–y &= 5 \end{align*}\]
The solution to a system of linear equations in two variables is any ordered pair that satisfies each equation independently. In this example, the ordered pair \((4,7)\) is the solution to the system of linear equations. We can verify the solution by substituting the values into each equation to see if the ordered pair satisfies both equations. Shortly we will investigate methods of finding such a solution if it exists.
\[\begin{align*} 2(4)+(7) &=15 \text{ True} \\ 3(4)−(7) &= 5 \text{ True} \end{align*}\]
In addition to considering the number of equations and variables, we can categorize systems of linear equations by the number of solutions. A consistent system of equations has at least one solution. A consistent system is considered to be an independent system if it has a single solution, such as the example we just explored. The two lines have different slopes and intersect at one point in the plane. A consistent system is considered to be a dependent system if the equations have the same slope and the same y -intercepts. In other words, the lines coincide so the equations represent the same line. Every point on the line represents a coordinate pair that satisfies the system. Thus, there are an infinite number of solutions.
Another type of system of linear equations is an inconsistent system , which is one in which the equations represent two parallel lines. The lines have the same slope and different y- intercepts. There are no points common to both lines; hence, there is no solution to the system.
There are three types of systems of linear equations in two variables, and three types of solutions.
- An independent system has exactly one solution pair \((x,y)\). The point where the two lines intersect is the only solution.
- An inconsistent system has no solution. Notice that the two lines are parallel and will never intersect.
- A dependent system has infinitely many solutions. The lines are coincident. They are the same line, so every coordinate pair on the line is a solution to both equations.
Figure \(\PageIndex{2}\) compares graphical representations of each type of system.
- Substitute the ordered pair into each equation in the system.
- Determine whether true statements result from the substitution in both equations; if so, the ordered pair is a solution.
Determine whether the ordered pair \((5,1)\) is a solution to the given system of equations.
\[\begin{align*} x+3y &= 8 \\ 2x−9 &= y \end{align*}\]
Solution
Substitute the ordered pair \((5,1)\) into both equations.
\[ \begin{align*} (5)+3(1) &= 8 \\ 8 &= 8 \text{ True} \\ 2(5)−9 &= (1) \\ 1 &= 1 \text{ True} \end{align*}\]
The ordered pair \((5,1)\) satisfies both equations, so it is the solution to the system.
Analysis
We can see the solution clearly by plotting the graph of each equation. Since the solution is an ordered pair that satisfies both equations, it is a point on both of the lines and thus the point of intersection of the two lines. See Figure \(\PageIndex{3}\).
Determine whether the ordered pair \((8,5)\) is a solution to the following system.
\[\begin{align*} 5x−4y &= 20 \\ 2x+1 &= 3y \end{align*}\]
- Answer
-
Not a solution.
Solving Systems of Equations by Graphing
There are multiple methods of solving systems of linear equations. For a system of linear equations in two variables, we can determine both the type of system and the solution by graphing the system of equations on the same set of axes.
Solve the following system of equations by graphing. Identify the type of system.
\[\begin{align*} 2x+y &= −8 \\ x−y &= −1 \end{align*}\]
Solution
Solve the first equation for \(y\).
\[\begin{align*} 2x+y &= −8 \\ y &= −2x−8 \end{align*}\]
Solve the second equation for \(y\).
\[\begin{align*} x−y &= −1 \\ y &= x+1 \end{align*}\]
Graph both equations on the same set of axes as in Figure \(\PageIndex{4}\).
The lines appear to intersect at the point \((−3,−2)\). We can check to make sure that this is the solution to the system by substituting the ordered pair into both equations.
\[\begin{align*} 2(−3)+(−2) &= −8 \\ −8 &= −8 \text{ True} \\ (−3)−(−2) &= −1 \\ −1 &= −1 \text{ True} \end{align*}\]
The solution to the system is the ordered pair \((−3,−2)\),so the system is independent.
Solve the following system of equations by graphing.
\[\begin{align*} 2x−5y &= −25 \\ −4x+5y &= 35 \end{align*}\]
- Answer
-
The solution to the system is the ordered pair \((−5,3)\).
Can graphing be used if the system is inconsistent or dependent?
Yes, in both cases we can still graph the system to determine the type of system and solution. If the two lines are parallel, the system has no solution and is inconsistent. If the two lines are identical, the system has infinite solutions and is a dependent system.
Solving Systems of Equations by Substitution
Solving a linear system in two variables by graphing works well when the solution consists of integer values, but if our solution contains decimals or fractions, it is not the most precise method. We will consider two more methods of solving a system of linear equations that are more precise than graphing. One such method is solving a system of equations by the substitution method , in which we solve one of the equations for one variable and then substitute the result into the second equation to solve for the second variable. Recall that we can solve for only one variable at a time, which is the reason the substitution method is both valuable and practical.
- Solve one of the two equations for one of the variables in terms of the other.
- Substitute the expression for this variable into the second equation, then solve for the remaining variable.
- Substitute that solution into either of the original equations to find the value of the first variable. If possible, write the solution as an ordered pair.
- Check the solution in both equations.
Solve the following system of equations by substitution.
\[\begin{align*} −x+y &= −5 \\ 2x−5y &= 1 \end{align*}\]
Solution
First, we will solve the first equation for \(y\).
\[\begin{align*} −x+y &=−5 \\ y &= x−5 \end{align*}\]
Now we can substitute the expression \(x−5\) for \(y\) in the second equation.
\[\begin{align*} 2x−5y &= 1 \\ 2x−5(x−5) &= 1 \\ 2x−5x+25 &= 1 \\ −3x &= −24 \\ x &= 8 \end{align*}\]
Now, we substitute \(x=8\) into the first equation and solve for \(y\).
\[\begin{align*} −(8)+y &= −5 \\ y &= 3 \end{align*}\]
Our solution is \((8,3)\).
Check the solution by substituting \((8,3)\) into both equations.
\[\begin{align*} −x+y &= −5 \\ −(8)+(3) &= −5 \text{ True} \\ 2x−5y &= 1 \\ 2(8)−5(3) &= 1 \text{ True} \end{align*}\]
Solve the following system of equations by substitution.
\[\begin{align*} x &= y+3 \\ 4 &= 3x−2y \end{align*}\]
- Answer
-
\((−2,−5)\)
Can the substitution method be used to solve any linear system in two variables?
Yes, but the method works best if one of the equations contains a coefficient of \(1\) or \(–1\) so that we do not have to deal with fractions.
Solving Systems of Equations in Two Variables by the Addition Method
A third method of solving systems of linear equations is the addition method. In this method, we add two terms with the same variable, but opposite coefficients, so that the sum is zero. Of course, not all systems are set up with the two terms of one variable having opposite coefficients. Often we must adjust one or both of the equations by multiplication so that one variable will be eliminated by addition.
- Write both equations with x - and y -variables on the left side of the equal sign and constants on the right.
- Write one equation above the other, lining up corresponding variables. If one of the variables in the top equation has the opposite coefficient of the same variable in the bottom equation, add the equations together, eliminating one variable. If not, use multiplication by a nonzero number so that one of the variables in the top equation has the opposite coefficient of the same variable in the bottom equation, then add the equations to eliminate the variable.
- Solve the resulting equation for the remaining variable.
- Substitute that value into one of the original equations and solve for the second variable.
- Check the solution by substituting the values into the other equation.
Solve the given system of equations by addition.
\[\begin{align*} x+2y &= −1 \\ −x+y &=3 \end{align*}\]
Solution
Both equations are already set equal to a constant. Notice that the coefficient of \(x\) in the second equation, \(–1\), is the opposite of the coefficient of \(x\) in the first equation, \(1\). We can add the two equations to eliminate \(x\) without needing to multiply by a constant.
\[\begin{align*} x+2y &= -1 \\ \underline{-x+y}& = \underline{3} \\ 3y&= 2 \\ \end{align*}\]
Now that we have eliminated \(x\), we can solve the resulting equation for \(y\).
\[\begin{align*} 3y &= 2 \\ y &=\dfrac{2}{3} \end{align*}\]
Then, we substitute this value for \(y\) into one of the original equations and solve for \(x\).
\[\begin{align*} −x+y &= 3 \\ −x+\dfrac{2}{3} &= 3 \\ −x &= 3−\dfrac{2}{3} \\ −x &= \dfrac{7}{3} \\ x &= −\dfrac{7}{3} \end{align*}\]
The solution to this system is \(\left(−\dfrac{7}{3},\dfrac{2}{3}\right)\).
Check the solution in the first equation.
\[\begin{align*} x+2y &= −1 \\ \left(−\dfrac{7}{3}\right)+2\left(\dfrac{2}{3}\right) &= \\ −\dfrac{7}{3}+\dfrac{4}{3} &= −\dfrac{3}{3} \\ −1 &= −1 \;\;\;\;\;\;\;\; \text{True} \end{align*}\]
Analysis
We gain an important perspective on systems of equations by looking at the graphical representation. See Figure \(\PageIndex{6}\) to find that the equations intersect at the solution. We do not need to ask whether there may be a second solution because observing the graph confirms that the system has exactly one solution.
Solve the given system of equations by the addition method.
\[\begin{align*} 3x+5y &= −11 \\ x−2y &= 11 \end{align*}\]
Solution
Adding these equations as presented will not eliminate a variable. However, we see that the first equation has \(3x\) in it and the second equation has \(x\). So if we multiply the second equation by \(−3\),the x -terms will add to zero.
\[\begin{align*} x−2y &= 11 \\ −3(x−2y) &=−3(11) \;\;\;\;\;\;\;\; \text{Multiply both sides by }−3. \\ −3x+6y &= −33 \;\;\;\;\;\;\;\;\; \text{Use the distributive property.} \end{align*}\]
Now, let’s add them.
\[\begin{align*} 3x+5y &= -11 \\ \underline{-3x+6y }& = \underline{-33} \\ 11y&= -44 \\ y&= -4 \end{align*}\]
For the last step, we substitute \(y=−4\) into one of the original equations and solve for \(x\).
\[\begin{align*} 3x+5y &= −11 \\ 3x+5(−4) &= −11 \\ 3x−20 &= −11 \\ 3x &= 9 \\ x &= 3 \end{align*}\]
Our solution is the ordered pair \((3,−4)\). See Figure \(\PageIndex{7}\). Check the solution in the original second equation.
\[\begin{align*} x−2y &= 11 \\ (3)−2(−4) &= 3+8 \\ &= 11 \;\;\;\;\;\;\;\;\;\; \text{True} \end{align*}\]
Solve the system of equations by addition.
\[\begin{align*} 2x−7y &= 2 \\ 3x+y &= −20 \end{align*}\]
- Answer
-
\((−6,−2)\)
Solve the given system of equations in two variables by addition.
\[\begin{align*} 2x+3y &= −16 \\ 5x−10y &= 30 \end{align*}\]
Solution
One equation has \(2x\) and the other has \(5x\). The least common multiple is \(10x\) so we will have to multiply both equations by a constant in order to eliminate one variable. Let’s eliminate \(x\) by multiplying the first equation by \(−5\) and the second equation by \(2\).
\[\begin{align*} −5(2x+3y) &= −5(−16) \\ −10x−15y &= 80 \\ 2(5x−10y) &= 2(30) \\ 10x−20y &= 60 \end{align*}\]
Then, we add the two equations together.
\[\begin{align*} -10x-15y &= 80 \\ \underline{10x-20y}& = \underline{60} \\ -35y&= 140 \\ y&= -4 \end{align*}\]
Substitute \(y=−4\) into the original first equation.
\[ \begin{align*} 2x+3(−4) &=−16 \\ 2x−12 &= −16 \\ 2x &= −4 \\ x &=−2 \end{align*}\]
The solution is \((−2,−4)\). Check it in the other equation.
\[\begin{align*} 5x−10y &= 30 \\ 5(−2)−10(−4) &= 30 \\ −10+40 &= 30 \\30 &=30 \end{align*}\]
See Figure \(\PageIndex{8}\).
Solve the given system of equations in two variables by addition.
\[ \begin{align*} \dfrac{x}{3}+\dfrac{y}{6} &= 3 \\ \dfrac{x}{2}−\dfrac{y}{4} &= 1 \end{align*}\]
Solution
First clear each equation of fractions by multiplying both sides of the equation by the least common denominator.
\[\begin{align*} 6\left(\dfrac{x}{3}+\dfrac{y}{6}\right) &= 6(3) \\ 2x+y &= 18 \\ 4\left(\dfrac{x}{2}−\dfrac{y}{4}\right) &= 4(1) \\ 2x−y &= 4 \end{align*}\]
Now multiply the second equation by \(−1\) so that we can eliminate the x -variable.
\[\begin{align*} −1(2x−y) &= −1(4) \\ −2x+y &= −4 \end{align*}\]
Add the two equations to eliminate the \(x\)-variable and solve the resulting equation.
\[\begin{align*} 2x+y &= 18 \\ −2x+y &= −4 \\ 2y &= 14 \\ y &=7 \end{align*}\]
Substitute \(y=7\) into the first equation.
\[\begin{align*} 2x+(7) &= 18 \\ 2x &= 11 \\ x &= \dfrac{11}{2} \\ &= 7.5 \end{align*}\]
The solution is \(\left(\dfrac{11}{2},7\right)\). Check it in the other equation.
\[\begin{align*} \dfrac{x}{2}−\dfrac{y}{4} &= 1 \\ \dfrac{\dfrac{11}{2}}{2}−\dfrac{7}{4} &=1 \\ \dfrac{11}{4}−\dfrac{7}{4} &=1 \\ \dfrac{4}{4} &=1 \end{align*}\]
Solve the system of equations by addition.
\[\begin{align*} 2x+3y &= 8 \\ 3x+5y &= 10 \end{align*}\]
- Answer
-
\((10,−4)\)
Identifying Inconsistent Systems of Equations Containing Two Variables
Now that we have several methods for solving systems of equations, we can use the methods to identify inconsistent systems. Recall that an inconsistent system consists of parallel lines that have the same slope but different y-intercepts. They will never intersect. When searching for a solution to an inconsistent system, we will come up with a false statement, such as \(12=0\).
Solve the following system of equations.
\[\begin{align*} x &= 9−2y \\ x+2y &= 13 \end{align*}\]
Solution
We can approach this problem in two ways. Because one equation is already solved for \(x\), the most obvious step is to use substitution.
\[\begin{align*} x+2y &= 13 \\ (9−2y)+2y &= 13 \\ 9+0y &= 13 \\ 9 &= 13 \end{align*}\]
Clearly, this statement is a contradiction because \(9≠13\). Therefore, the system has no solution.
The second approach would be to first manipulate the equations so that they are both in slope-intercept form. We manipulate the first equation as follows.
\[\begin{align*} x &= 9−2y \\ 2y &= −x+9 \\ y &= −\dfrac{1}{2}x+\dfrac{9}{2} \end{align*}\]
We then convert the second equation expressed to slope-intercept form.
\[\begin{align*} x+2y &= 13 \\ 2y &= −x+13 \\ y &= −\dfrac{1}{2}x+\dfrac{13}{2} \end{align*}\]
Comparing the equations, we see that they have the same slope but different \(y\)-intercepts. Therefore, the lines are parallel and do not intersect.
\[\begin{align*} y &= −\dfrac{1}{2}x+\dfrac{9}{2} \\ y &= −\dfrac{1}{2}x+\dfrac{13}{2} \end{align*}\]
Analysis
Writing the equations in slope-intercept form confirms that the system is inconsistent because all lines will intersect eventually unless they are parallel. Parallel lines will never intersect; thus, the two lines have no points in common. The graphs of the equations in this example are shown in Figure \(\PageIndex{9}\).
Solve the following system of equations in two variables.
\[\begin{align*} 2y−2x &= 2 \\ 2y−2x &= 6 \end{align*}\]
- Answer
-
No solution. It is an inconsistent system.
Expressing the Solution of a System of Dependent Equations Containing Two Variables
Recall that a dependent system of equations in two variables is a system in which the two equations represent the same line. Dependent systems have an infinite number of solutions because all of the points on one line are also on the other line. After using substitution or addition, the resulting equation will be an identity, such as \(0=0\).
Find a solution to the system of equations using the addition method .
\[\begin{align*} x+3y &= 2 \\ 3x+9y &= 6 \end{align*}\]
Solution
With the addition method, we want to eliminate one of the variables by adding the equations. In this case, let’s focus on eliminating \(x\). If we multiply both sides of the first equation by \(−3\), then we will be able to eliminate the x-variable.
\[\begin{align*} x+3y &= 2 \\ (−3)(x+3y) &= (−3)(2) \\ −3x−9y &= −6 \end{align*}\]
Now add the equations.
\[\begin{align*} -3x-9y &= -6 \\ \underline{+\space 3x+9y}& = \underline{6} \\ 0&= 0 \\ \end{align*}\]
We can see that there will be an infinite number of solutions that satisfy both equations.
Analysis
If we rewrote both equations in the slope-intercept form, we might know what the solution would look like before adding. Let’s look at what happens when we convert the system to slope-intercept form.
\[ \begin{align*} x+3y &= 2 \\ 3y &= −x+2 \\ y &= −\dfrac{1}{3}x+\dfrac{2}{3} \\ 3x+9y &= 6 \\ 9y &=−3x+6 \\ y &= −\dfrac{3}{9}x+\dfrac{6}{9} \\ y &= −\dfrac{1}{3}x+\dfrac{2}{3} \end{align*}\]
See Figure \(\PageIndex{10}\). Notice the results are the same. The general solution to the system is \(\left(x, −\dfrac{1}{3}x+\dfrac{2}{3}\right)\).
Solve the following system of equations in two variables.
\[\begin{align*} y−2x &= 5 \\ −3y+6x &= −15 \end{align*}\]
- Answer
-
The system is dependent so there are infinite solutions of the form \((x,2x+5)\).
Using Systems of Equations to Investigate Profits
Using what we have learned about systems of equations, we can return to the skateboard manufacturing problem at the beginning of the section. The skateboard manufacturer’s revenue function is the function used to calculate the amount of money that comes into the business. It can be represented by the equation \(R=xp\), where \(x\)=quantity and \(p\)=price. The revenue function is shown in orange in Figure \(\PageIndex{11}\).
The cost function is the function used to calculate the costs of doing business. It includes fixed costs, such as rent and salaries, and variable costs, such as utilities. The cost function is shown in blue in Figure \(\PageIndex{11}\). The \(x\)-axis represents quantity in hundreds of units. The \(y\)-axis represents either cost or revenue in hundreds of dollars.
The point at which the two lines intersect is called the break-even point. We can see from the graph that if \(700\) units are produced, the cost is \($3,300\) and the revenue is also \($3,300\). In other words, the company breaks even if they produce and sell \(700\) units. They neither make money nor lose money.
The shaded region to the right of the break-even point represents quantities for which the company makes a profit. The shaded region to the left represents quantities for which the company suffers a loss. The profit function is the revenue function minus the cost function, written as \(P(x)=R(x)−C(x)\). Clearly, knowing the quantity for which the cost equals the revenue is of great importance to businesses.
Given the cost function \(C(x)=0.85x+35,000\) and the revenue function \(R(x)=1.55x\), find the break-even point and the profit function.
Solution
Write the system of equations using \(y\) to replace function notation.
\[\begin{align*} y &= 0.85x+35,000 \\ y &= 1.55x \end{align*}\]
Substitute the expression \(0.85x+35,000\) from the first equation into the second equation and solve for \(x\).
\[\begin{align*} 0.85x+35,000 &= 1.55x \\ 35,000 &= 0.7x \\ 50,000 &= x \end{align*}\]
Then, we substitute \(x=50,000\) into either the cost function or the revenue function.
\(1.55(50,000)=77,500\)
The break-even point is \((50,000,77,500)\).
The profit function is found using the formula \(P(x)=R(x)−C(x)\).
\[\begin{align*} P(x) &= 1.55x−(0.85x+35,000) \\ &=0.7x−35,000 \end{align*}\]
The profit function is \(P(x)=0.7x−35,000\).
Analysis
The cost to produce \(50,000\) units is \($77,500\), and the revenue from the sales of \(50,000\) units is also \($77,500\). To make a profit, the business must produce and sell more than \(50,000\) units. See Figure \(\PageIndex{12}\).
We see from the graph in Figure \(\PageIndex{13}\) that the profit function has a negative value until \(x=50,000\), when the graph crosses the \(x\)-axis. Then, the graph emerges into positive \(y\)-values and continues on this path as the profit function is a straight line. This illustrates that the break-even point for businesses occurs when the profit function is \(0\). The area to the left of the break-even point represents operating at a loss.
The cost of a ticket to the circus is \($25.00\) for children and \($50.00\) for adults. On a certain day, attendance at the circus is \(2,000\) and the total gate revenue is \($70,000\). How many children and how many adults bought tickets?
Solution
Let \(c\) = the number of children and \(a\) = the number of adults in attendance.
The total number of people is \(2,000\). We can use this to write an equation for the number of people at the circus that day.
\(c+a=2,000\)
The revenue from all children can be found by multiplying \($25.00\) by the number of children, \(25c\). The revenue from all adults can be found by multiplying \($50.00\) by the number of adults, \(50a\). The total revenue is \($70,000\). We can use this to write an equation for the revenue.
\(25c+50a=70,000\)
We now have a system of linear equations in two variables.
\(c+a=2,000\)
\(25c+50a=70,000\)
In the first equation, the coefficient of both variables is \(1\). We can quickly solve the first equation for either \(c\) or \(a\). We will solve for \(a\).
\[\begin{align*} c+a &= 2,000 \\ a &= 2,000−c \end{align*}\]
Substitute the expression \(2,000−c\) in the second equation fora a and solve for \(c\).
\[\begin{align*} 25c+50(2,000−c) &= 70,000 \\ 25c+100,000−50c &= 70,000 \\ −25c &= −30,000 \\ c &= 1,200 \end{align*}\]
Substitute \(c=1,200\) into the first equation to solve for \(a\).
\[\begin{align*} 1,200+a &= 2,000 \\ a &= 800 \end{align*}\]
We find that \(1,200\) children and \(800\) adults bought tickets to the circus that day.
Meal tickets at the circus cost \($4.00\) for children and \($12.00\) for adults. If \(1,650\) meal tickets were bought for a total of, \($14,200\), how many children and how many adults bought meal tickets?
- Answer
-
\(700\) children, \(950\) adults
Access these online resources for additional instruction and practice with systems of linear equations.
Key Concepts
- A system of linear equations consists of two or more equations made up of two or more variables such that all equations in the system are considered simultaneously.
- The solution to a system of linear equations in two variables is any ordered pair that satisfies each equation independently. See Example \(\PageIndex{1}\).
- Systems of equations are classified as independent with one solution, dependent with an infinite number of solutions, or inconsistent with no solution.
- One method of solving a system of linear equations in two variables is by graphing. In this method, we graph the equations on the same set of axes. See Example \(\PageIndex{2}\).
- Another method of solving a system of linear equations is by substitution. In this method, we solve for one variable in one equation and substitute the result into the second equation. See Example \(\PageIndex{3}\).
- A third method of solving a system of linear equations is by addition, in which we can eliminate a variable by adding opposite coefficients of corresponding variables. See Example \(\PageIndex{4}\).
- It is often necessary to multiply one or both equations by a constant to facilitate elimination of a variable when adding the two equations together. See Example \(\PageIndex{5}\), Example \(\PageIndex{6}\), and Example \(\PageIndex{7}\).
- Either method of solving a system of equations results in a false statement for inconsistent systems because they are made up of parallel lines that never intersect. See Example \(\PageIndex{8}\).
- The solution to a system of dependent equations will always be true because both equations describe the same line. See Example \(\PageIndex{9}\).
- Systems of equations can be used to solve real-world problems that involve more than one variable, such as those relating to revenue, cost, and profit. See Example \(\PageIndex{10}\) and Example \(\PageIndex{11}\).