5.1.7: A Final Thought
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One of the most widely recognized cultural distinctions in psychology today is the difference between individualistic, Western cultures and collectivistic, Eastern cultures. In Western societies, such as the United States of America, the individual not only has the freedom to seek purely personal advancement, it is expected of them. In contrast, the individual in countries such as China is expected to subordinate their own desires and ambitions for the good of the family and their community. With regard to a broad view of the African personality, we find a middle ground. There is significant individual freedom, but individuals are expected to serve their family and community. As a result, the individual also benefits from the overall success of the family and community. Thus, there is an ongoing interplay between the value of the individual and the values of family and community.
When this system works to its best potential, the results are people who flourish and can be proud of themselves. In the words of Dr. J. A. Sofola:
…the philosophy, the world-view, values and thought-patterns that form the ingredients or the building-blocks of the African Personality are live-and-let-live; the emphasis on wholesome human relations; the belief of the universality of man and communality of the people in the community; the historic sense of the unity of the human society as consisting of the ancestors, the living and the future generations yet unborn; spiritual attitude to life and attachment to communal life with communal responsibilities; a keen sense of rhythm; the conception of man as one roaming spirit in the chain of spirits in the universe…This is the personality which in its expression of an inward peace and stillness maintains an external composure and gait, head and chin raised high, and with deliberate, calculated dignified steps proclaims to the world: “Black is beautiful” and “I am black and proud of being so.” (pp. 143-144; Sofola, 1973)


16.3E: Cold-Seep Ecosystems
- Outline the organisms that live in cold-seep ecosystems
A cold seep (sometimes called a cold vent) is an area of the ocean floor where hydrogen sulfide, methane, and other hydrocarbon-rich fluid seepage occurs, often in the form of a brine pool. “Cold” does not mean temperature of seepage is lower than surrounding sea water. Actually, its temperature is often slightly higher. Cold seeps constitute a biome supporting several endemic species.
Cold seeps develop unique topography over time, where reactions between methane and seawater create carbonate rock formations and reefs. These reactions may also be dependent on bacterial activity. Ikaite, a hydrous calcium carbonate, can be associated with oxidizing methane at cold seeps.
Types of cold seeps can be distinguished according to the depth, as shallow cold seeps and deep cold seeps. Cold seeps can also be distinguished in detail, as follows: oil/gas seeps, gas seeps, methane seeps, gas hydrate seeps, brine seeps, are forming brine pools, pockmarks and mud volcanos.
Organisms living in cold seeps are known as extremophiles. Biological research in cold seeps and hydrothermal vents has been mostly focused on the microbiology and the prominent chemosynthetic macro-invertebrates. Much less research has been done on the smaller benthic fraction at the size of the meiofauna (<1 mm).
Community composition’s orderly shift from one set of species to another is called ecological succession. The first type of organism to take advantage of this deep-sea energy source is bacteria. Aggregating into bacterial mats at cold seeps, these bacteria metabolize methane and hydrogen sulfide (another gas that emerges from seeps) for energy. This process of obtaining energy from chemicals is known as chemosynthesis.
During this initial stage, when methane is relatively abundant, dense mussel beds also form near the cold seep. Mostly composed of species in the genus Bathymodiolus, these mussels do not directly consume food. Instead, they are nourished by symbiotic bacteria that also produce energy from methane, similar to their relatives that form mats. Chemosynthetic bivalves are prominent constituents of the fauna of cold seeps and are represented in that setting by five families: Solemyidae, Lucinidae, Vesicomyidae, Thyasiridae, and Mytilidae.
This microbial activity produces calcium carbonate (CaCO3), which is deposited on the seafloor and forms a layer of rock. During a period lasting up to several decades, these rock formations attract siboglinid tubeworms, which settle and grow along with the mussels. Like the mussels, tubeworms rely on chemosynthetic bacteria (in this case, a type that needs hydrogen sulfide instead of methane) for survival. True to any symbiotic relationship, a tubeworm also provides for their bacteria by appropriating hydrogen sulfide from the environment. The sulfide not only comes from the water, but is also mined from the sediment through an extensive “root” system a tubeworm “bush” establishes in the hard, carbonate substrate. A tubeworm bush can contain hundreds of individual worms, which can grow a meter or more above the sediment.
Cold seeps do not last indefinitely.  At this stage, tubeworms become the dominant organism in a seep community. As long as there is some sulfide in the sediment, the sulfide-mining tubeworms can persist. Individuals of one tubeworm species Lamellibrachia luymesi have been estimated to live for over 250 years in such conditions.
Key Points
- Cold seeps develop unique topography over time, where reactions between methane and seawater create carbonate rock formations and reefs.
- Types of cold seeps can be distinguished according to the depth, as shallow cold seeps and deep cold seeps.
- Organisms living in cold seeps are known as extremophiles.
Key Terms
- cold seep : A cold seep (sometimes called a cold vent) is an area of the ocean floor where hydrogen sulfide, methane, and other hydrocarbon-rich fluid seepage occurs, often in the form of a brine pool. “Cold” does not mean temperature of seepage is lower than surrounding sea water. Actually, its temperature is often slightly higher.
- topography : A detailed graphic representation of the surface features of a place or object.
- extremophiles : An extremophile (from Latin extremus, meaning “extreme,” and Greek philiā (φ), meaning “love”) is an organism that thrives in physically or geochemically extreme conditions that are detrimental to most life on earth.


3.3: Radical Equations
By the end of this section, you will be able to:
- Solve radical equations involving one or two radicals.
- Solve equations involving rational exponents.
- Solve equations that are quadratic-in-form involving radicals or rational exponents.
- Use radicals in applications
As is the theme in College Algebra, we begin this section with the assumption that you already have exposure to solving simple radical equations. We still spend a few moments recalling old techniques, but quickly jump into solving more complex radical equations.
Solving Radical Equations Containing One Radical
We start our trip down memory lane by recalling an old definition.
An equation in which a variable is in the radicand of a radical expression is called a radical equation .
When solving these equations, our goal is to isolate the radical. At that point, our strategy will be to raise both sides of the equation to the power of the index. This will, in turn, eliminate the radical.
Solving radical equations containing an even index by raising both sides to the power of the index may introduce an algebraic solution that would not be a solution to the original radical equation. Again, we call this an extraneous solution as we did when we solved rational equations.
In the next example, we will see how to solve a radical equation. Our strategy is based on raising a radical with index \(n\) to the \(n^{\text{th}}\) power. This will eliminate the radical.
Solve: \(\sqrt{5 n-4}-9=0\).
- Solution
-
\[ \begin{array}{rrclcl}
& \sqrt{5n - 4} - 9 & = & 0 & & \\[6pt]
\implies & \sqrt{5n - 4} & = & 9 & \quad & \left( \text{isolating the radical by adding }9\text{ to both sides} \right) \\[6pt]
\implies & \left(\sqrt{5n - 4}\right)^2 & = & 9^2 & \quad & \left( \text{squaring both sides. }\textbf{Warning: }\text{This forces us to check our final solution} \right) \\[6pt]
\implies & 5n - 4 & = & 81 & \quad & \left( \text{simplifying} \right) \\[6pt]
\implies & 5n & = & 85 & \quad & \left( \text{adding }4\text{ to both sides} \right) \\[6pt]
\implies & n & = & 17 & \quad & \left( \text{dividing both sides by 5} \right) \\[6pt]
\end{array} \nonumber \]Now we must to check the solution. When doing so, you need to go back to the original equation to check your answer.\[ \sqrt{5(17) - 4} - 9 = \sqrt{85 - 4} - 9 = \sqrt{81} - 9 = 9 - 9 = 0. \checkmark \nonumber \]Thus, \( n = 17 \) is the solution to the equation.
Solve: \(\sqrt{3 m+2}-5=0\).
- Answer
-
\(m=\frac{23}{3}\)
It is important to remember the following advice:
Whenever you raise both sides of an equation to an even power, you must check your final solutions.
In truth, the reason why we (as mathematicians) ask you to do this is to catch any inconsistencies that might have slipped past you. For example, suppose we were asked to solve\[ \sqrt{3x - 1} - 7 = -9. \nonumber \]We would start by isolating the radical to get\[ \sqrt{3x - 1} = -2. \nonumber \]If you are paying attention (and not asleep at the "mathematical wheel"), you would notice that you have an even-indexed radical equaling a negative number - this cannot happen! At this point, you would stop the solution process and state that the solution doesn't exist (some people use the notation "DNE" for "does not exist").
However, if you didn't notice the negative on the right side, you would continue to solve by squaring both sides to get\[ 3x - 1 = 4 \implies 3x = 5 \implies x = \dfrac{5}{3}. \nonumber \]As long as we develop the habit to check our solutions whenever we square both sides of an equation, we have a second safety net.\[ \sqrt{3 \left( \dfrac{5}{3} \right) - 1} - 7 = \sqrt{5 - 7} - 7 = \sqrt{-2} - 7 \nonumber \]which is not a real number. As such, it cannot become \( -9 \).
I often try to get my students to stop at each step of the solution process to think about whether or not the statement before them makes sense, but checking your work at the end is your second opportunity to catch the logical flaw.
Solve: \(\sqrt{9 k-2}+1=0\).
- Solution
-
\[ \begin{array}{rrclcl}
& \sqrt{9k - 2} + 1 & = & 0 & & \\[6pt]
\implies & \sqrt{9k - 2} & = & -1 & \quad & \left( \text{isolating the radical by subtracting }1\text{ from both sides} \right) \\[6pt]
\end{array} \nonumber \]At this point, if we were paying attention, we could stop because a square root is never equal to \( -1 \) (at least, not over the real numbers).
Having recognized the square root equaling a negative number in the last step of Example \( \PageIndex{ 2 } \) saved us time. Had we not seen that issue, we would have continued on to get\[ \begin{array}{rrclcl}
& \sqrt{9k - 2} + 1 & = & 0 & & \\[6pt]
\implies & \sqrt{9k - 2} & = & -1 & \quad & \left( \text{isolating the radical by subtracting }1\text{ from both sides} \right) \\[6pt]
\implies & 9k - 2 & = & 1 & \quad & \left( \text{squaring both sides} \right) \\[6pt]
\implies & 9k & = & 3 & \quad & \left( \text{adding }2\text{ to both sides} \right) \\[6pt]
\implies & k & = & \dfrac{1}{3} & \quad & \left( \text{dividing both sides by }9 \right) \\[6pt]
\end{array} \nonumber \]Because we squared both sides, we go back to the original equation and check our answer.\[ \sqrt{9\left( \dfrac{1}{3} \right) - 2} + 1 = \sqrt{3 - 2} + 1 = \sqrt{1} + 1 = 1 + 1 = 2 \neq 0. \nonumber \]This is a
contradiction
. Hence, \( k = \frac{1}{3} \) is an extraneous solution.
Solve: \(\sqrt{2 r-3}+5=0\).
- Answer
-
no solution
Solve: \(\sqrt{p-1}+1=p\).
- Solution
-
\[ \begin{array}{rrclcl}
& \sqrt{p - 1} + 1 & = & p & & \\[6pt]
\implies & \sqrt{p - 1} & = & p - 1 & \quad & \left( \text{isolating the radical by subtracting }1\text{ from both sides} \right) \\[6pt]
\implies & \left(\sqrt{p - 1}\right)^2 & = & \left(p - 1\right)^2 & \quad & \left( \text{squaring both sides} \right) \\[6pt]
\implies & p - 1 & = & p^2 - 2p + 1 & \quad & \left( \text{simplifying and distributing} \right) \\[6pt]
\implies & 0 & = & p^2 - 3p + 2 & \quad & \left( \text{subtracting }p\text{ and adding }1\text{ to both sides} \right) \\[6pt]
\implies & 0 & = & (p - 2)(p - 1) & \quad & \left( \text{factoring} \right) \\[6pt]
\end{array} \nonumber \]Therefore, our candidate solutions are \( p = 2 \) or \( p = 1 \). However, because we squared both sides, we need to check these results (using the original equation).\[ \begin{array}{rrcl|rrcl}
p = 2 & & & & p = 1 & & & \\[6pt]
\implies & \sqrt{2 - 1} + 1 & \overset{\text{?}}{=} & 2 & \implies & \sqrt{1 - 1} + 1 & \overset{\text{?}}{=} & 1 \\[6pt]
\implies & \sqrt{1} + 1 & \overset{\text{?}}{=} & 2 & \implies & \sqrt{0} + 1 & \overset{\text{?}}{=} & 1 \\[6pt]
\implies & 1 + 1 & \overset{\text{?}}{=} & 2 & \implies & 0 + 1 & \overset{\text{?}}{=} & 1 \\[6pt]
\implies & 2 & = & 2 \checkmark & \implies & 1 & = & 1 \checkmark \\[6pt]
\end{array} \nonumber \]Therefore, the solutions are \(p=1\) and \(p=2\).
Solve: \(\sqrt{x-2}+2=x\).
- Answer
-
\(x=2, x=3\)
It should be obvious that when the index of the radical is \(n\), we eventually raise both sides of the radical equation to the \( n^{\text{th}} \) power.
Solve: \(\sqrt[3]{5 x+1}+8=4\).
- Solution
-
\[ \begin{array}{rrclcl}
& \sqrt[3]{5x + 1} + 8 & = & 4 & & \\[6pt]
\implies & \sqrt[3]{5x + 1} & = & -4 & \quad & \left( \text{isolating the radical by subtracting }4\text{ from both sides} \right) \\[6pt]
\implies & \left(\sqrt[3]{5x + 1}\right)^3 & = & \left(-4\right)^3 & \quad & \left( \text{raising both sides to the third power} \right) \\[6pt]
\implies & 5x + 1 & = & -64 & \quad & \left( \text{simplifying} \right) \\[6pt]
\implies & 5x & = & -65 & \quad & \left( \text{subtracting }1\text{ from both sides} \right) \\[6pt]
\implies & x & = & -13 & \quad & \left( \text{dividing both sides by }5 \right) \\[6pt]
\end{array} \nonumber \]
It's always a good idea to check your answers when solving equations in Mathematics; however, it's only truly required (at least for now) when we have raised both sides of an equation to an even power (e.g., when solving radical equations) or when we have multiplied both sides of an equation by an expression containing a variable (e.g., when solving a rational equation). This is why we did not bother checking our solution in Example \( \PageIndex{ 4 } \). We leave it to the reader to check that \( x = -13 \) works.
Solve: \( \sqrt[3]{4 x-3}+8=5\)
- Answer
-
\(x=-6\)
Solving Equations Containing Rational Exponents
Sometimes an equation will contain rational exponents instead of a radical. We use the same techniques to solve the equation as when we have a radical equation. Since \( a^{m/n} = \sqrt[n]{a^m} \), we first raise each side of the equation to the power of the denominator of the rational exponent. Once that is done, we take the \( m^{\text{th}} \) root of both sides of the equation (or raise both sides to the \( \frac{1}{m}^{\text{th}} \) power).
Solve: \((3 x-2)^{1/4}+3=5\).
- Solution
-
\[ \begin{array}{rrclcl}
& (3 x-2)^{1/4}+3 & = & 5 & & \\[6pt]
\implies & (3 x-2)^{1/4} & = & -2 & \quad & \left( \text{subtracting }3\text{ from both sides} \right) \\[6pt]
\implies & \left((3 x-2)^{1/4}\right)^4 & = & \left(-2\right)^4 & \quad & \left( \text{raising both sides to the fourth power} \right) \\[6pt]
\implies & (3 x-2)^{4/4} & = & 16 & \quad & \left( \text{Laws of Exponents} \right) \\[6pt]
\implies & (3 x-2)^{1} & = & 16 & \quad & \left( \text{simplifying} \right) \\[6pt]
\implies & 3 x-2 & = & 16 & \quad & \left( \text{simplifying} \right) \\[6pt]
\implies & 3 x & = & 18 & \quad & \left( \text{adding }2\text{ to both sides} \right) \\[6pt]
\implies & x & = & 6 & \quad & \left( \text{dividing both sides by }3 \right) \\[6pt]
\end{array} \nonumber \]Since we raised both sides to an even power, we need to check for extraneous solutions. Going back to the original equation and substituting \( x = 6 \), we get\[ \left( 3(6) - 2 \right)^{1/4} + 3 = \left( 18 - 2 \right)^{1/4} + 3 = (16)^{1/4} + 3 = \sqrt[4]{16} + 3 = 2 + 3 = 5 \, \checkmark. \nonumber \]
Solve: \((9 x+9)^{1/4}-2=1\)
- Answer
-
\(x=8\)
Care must be taken when solving equations containing rational exponents. We must remember all of the techniques we previously learned. For example, in the equation\[ a^{4/7} = 1, \nonumber \]we start by raising both sides to the \( 7^{\text{th}} \) power to get\[ a^4 = 1^7 = 1. \nonumber \]However, we now have the variable \( a \) raised to an even power. Extraction of Roots tells us that \( a = \pm 1 \). This is important!
Solve: \( 3\left( \frac{1}{2}x + 1 \right)^{2/3} - 5 = 7 \)
- Solution
-
\[ \begin{array}{rrclcl}
& 3\left( \dfrac{1}{2}x + 1 \right)^{2/3} - 5 & = & 7 & & \\[6pt]
\implies & 3\left( \dfrac{1}{2}x + 1 \right)^{2/3} & = & 12 & \quad & \left( \text{begin isolating the rational exponent expression by adding }5\text{ to both sides} \right) \\[6pt]
\implies & \left( \dfrac{1}{2}x + 1 \right)^{2/3} & = & 4 & \quad & \left( \text{continue isolating the rational exponent expression by dividing both sides by }3 \right) \\[6pt]
\implies & \left(\left( \dfrac{1}{2}x + 1 \right)^{2/3}\right)^3 & = & \left(4\right)^3 & \quad & \left( \text{raise both sides to the third power} \right) \\[6pt]
\implies & \left( \dfrac{1}{2}x + 1 \right)^{6/3} & = & 64 & \quad & \left( \text{Laws of Exponents} \right) \\[6pt]
\implies & \left( \dfrac{1}{2}x + 1 \right)^{2} & = & 64 & \quad & \left( \text{simplifying} \right) \\[6pt]
\implies & \dfrac{1}{2}x + 1 & = & \pm 8 & \quad & \left( \text{Extraction of Roots} \right) \\[6pt]
\implies & \dfrac{1}{2}x & = & -1 \pm 8 & \quad & \left( \text{subtracting }1\text{ from both sides} \right) \\[6pt]
\implies & x & = & -2 \pm 16 & \quad & \left( \text{multiplying both sides by }2 \right) \\[6pt]
\end{array} \nonumber \]Thus, the solutions are \( x = -2 - 16 = -18 \) and \( x = -2 + 16 = 14 \).
Solve: \( 6\left( 3x - 5 \right)^{3/2} - 4 = 2 \)
- Answer
-
\( x = 2 \)
Solving Radical Equations with Two Radicals
If we have a radical equation containing two radicals, we start out by isolating one of them. It often works out easiest to isolate the more complicated radical first. In the next example, when one radical is isolated, the second radical is also isolated.
Solve: \(\sqrt[3]{4 x-3} - \sqrt[3]{3 x+2} = 0\).
- Solution
-
\[ \begin{array}{rrclcl}
& \sqrt[3]{4 x-3} - \sqrt[3]{3 x+2} & = & 0 & & \\[6pt]
\implies & \sqrt[3]{4 x-3} & = & \sqrt[3]{3 x+2} & \quad & \left( \text{isolating one of the radicals (which actually isolates the other in this example)} \right) \\[6pt]
\implies & \left(\sqrt[3]{4 x-3}\right)^3 & = & \left(\sqrt[3]{3 x+2}\right)^3 & \quad & \left( \text{raising both sides to the third power} \right) \\[6pt]
\implies & 4 x-3 & = & 3 x+2 & \quad & \left( \text{simplifying} \right) \\[6pt]
\implies & x & = & 5 & \quad & \left( \text{subtracting }3x\text{ and adding }3\text{ to both sides} \right) \\[6pt]
\end{array} \nonumber \]The solution is \(x=5\). We leave it to you to show that \(5\) checks.
Solve: \(\sqrt[3]{7 x+1}=\sqrt[3]{2 x-5}\).
- Answer
-
\(x=-\frac{6}{5}\)
Sometimes after raising both sides of an equation to a power, we still have a variable inside a radical. When that happens, we "rinse and repeat" by isolating the remaining radical and raising both sides of the equation to the power of the index again.
Solve: \(\sqrt{m}+1=\sqrt{m+9}\).
- Solution
-
\[ \begin{array}{rrclcl}
& \sqrt{m} + 1 & = & \sqrt{m + 9} & & \\[6pt]
\implies & \left( \sqrt{m} + 1 \right)^2 & = & \left( \sqrt{m + 9} \right)^2 & \quad & \left( \text{since a radical was already isolated, square both sides} \right) \\[6pt]
\implies & \left( \sqrt{m} + 1 \right)\left( \sqrt{m} + 1 \right) & = & m + 9 & \quad & \left( \text{simplifying} \right) \\[6pt]
\implies & m + 2 \sqrt{m} + 1 & = & m + 9 & \quad & \left( \text{distributing} \right) \\[6pt]
\implies & 2 \sqrt{m} & = & 8 & \quad & \left( \text{begin isolating the remaining radical by subtracting both }m\text{ and }1\text{ from both sides} \right) \\[6pt]
\implies & \sqrt{m} & = & 4 & \quad & \left( \text{dividing both sides by }2 \right) \\[6pt]
\implies & \left(\sqrt{m}\right)^2 & = & 4^2 & \quad & \left( \text{squaring both sides} \right) \\[6pt]
\implies & m & = & 16 & \quad & \left( \text{simplifying} \right) \\[6pt]
\end{array} \nonumber \]Since we squared both sides during our solution process, we need to check that our solution works in the original equation.\[ \begin{array}{rrcl}
& \sqrt{16} + 1 & \overset{\text{?}}{=} & \sqrt{16 + 9} \\[6pt]
\implies & 4 + 1 & \overset{\text{?}}{=} & \sqrt{25} \\[6pt]
\implies & 5 & = & 5 \, \checkmark \\[6pt]
\end{array} \nonumber \]Therefore, the solution is \( m = 16 \).
Solve: \(3-\sqrt{x}=\sqrt{x-3}\).
- Answer
-
\(x=4\)
We summarize the steps here. We have adjusted our previous steps to include more than one radical in the equation This procedure will now work for any radical equation.
- Isolate one of the radical terms on one side of the equation.
- Raise both sides of the equation to the power of the index.
-
Are there any more radicals?
If yes, repeat Step 1 and Step 2 again.
If no, solve the new equation. - Check the answer in the original equation.
Be careful as you square binomials (as in in the next example). Remember the pattern in \((a+b)^{2}=a^{2}+2 a b+b^{2}\) or \((a-b)^{2}=a^{2}-2 a b+b^{2}\).
Solve: \(\sqrt{q-2}+3=\sqrt{4 q+1}\).
- Solution
-
\[ \begin{array}{rrclcl}
& \sqrt{q-2} + 3 & = & \sqrt{4q+1} & & \\[6pt]
\implies & \left( \sqrt{q-2}+3 \right)^2 & = & \left( \sqrt{4 q+1} \right)^2 & \quad & \left( \text{squaring both sides since one radical is already isolated} \right) \\[6pt]
\implies & \left( \sqrt{q-2}+3 \right)\left( \sqrt{q-2}+3 \right) & = & 4 q+1 & \quad & \left( \text{simplifying} \right) \\[6pt]
\implies & q - 2 + 6 \sqrt{q-2} + 9 & = & 4 q+1 & \quad & \left( \text{distributing} \right) \\[6pt]
\implies & q + 7 + 6 \sqrt{q-2} & = & 4 q+1 & \quad & \left( \text{combining like terms} \right) \\[6pt]
\implies & 6 \sqrt{q-2} & = & 3 q - 6 & \quad & \left( \text{begin isolating the remaining radical by subtracting }q\text{ and }7\text{ from both sides} \right) \\[6pt]
\implies & 2 \sqrt{q-2} & = & q - 2 & \quad & \left( \text{noticing that all coefficients are share a common divisor, we divide both sides by }3 \right) \\[6pt]
\implies & \left(2 \sqrt{q-2}\right)^2 & = & \left(q - 2\right)^2 & \quad & \left( \text{instead of dividing by }2\text{ (which would introduce fractions), we square both sides} \right) \\[6pt]
\implies & 2^2 \left(\sqrt{q-2}\right)^2 & = & \left(q - 2\right)\left(q - 2\right) & \quad & \left( \text{Laws of Exponents} \right) \\[6pt]
\implies & 4 \left(q-2\right) & = & q^2 - 4q + 4 & \quad & \left( \text{simplifying and distributing} \right) \\[6pt]
\implies & 4 q - 8 & = & q^2 - 4q + 4 & \quad & \left( \text{distributing} \right) \\[6pt]
\implies & 0 & = & q^2 - 8q + 12 & \quad & \left( \text{subtracting }4q\text{ and adding }8\text{ to both sides} \right) \\[6pt]
\implies & 0 & = & \left(q - 6\right)\left(q - 2\right) & \quad & \left( \text{factoring the trinomial} \right) \\[6pt]
\end{array} \nonumber \]Therefore, our candidate solutions are \( q = 6 \) and \( q = 2 \). We leave it to the reader to check that neither is extraneous.
Solve: \(\sqrt{x-1}+2=\sqrt{2 x+6}\)
- Answer
-
\(x=5\)
Focusing on Calculus - Revisiting Solving Equations that are Quadratic-in-Form
Equations involving rational exponents are fertile territory to run into equations that are quadratic-in-form. Not only must you recognize their form, but you also must watch out for all the pitfalls mentioned throughout this section concerning checking your solutions.
Solve: \( x - 2 \sqrt{x} - 8 = 0 \).
- Solution
-
This is a radical equation that can be written in quadratic form. If we let \( u = \sqrt{x} \), then \( u^2 = \left( \sqrt{x} \right)^2 = x \) and we can write\[ \begin{array}{rrclcl}
& x - 2\sqrt{x} - 8 & = & 0 & & \\[6pt]
\implies & u^2 - 2u - 8 & = & 0 & \quad & \left( \text{substituting} \right) \\[6pt]
\implies & (u - 4)(u + 2) & = & 0 & \quad & \left( \text{factoring the trinomial} \right) \\[6pt]
\end{array} \nonumber \]Therefore, \( u = 4 \) or \( u = -2 \). Since \( u = \sqrt{x} \), we resubstitute to get \( \sqrt{x} = 4 \) or \( \sqrt{x} = -2 \); however, square roots are never negative. Hence, we throw out that possibility and focus on \( \sqrt{x} = 4 \implies x = 16 \). Checking this solution in the original equation, we get\[ 16 - 2 \sqrt{16} - 8 = 16 - 2(4) - 8 = 16 - 8 - 8 = 0. \, \checkmark \nonumber \]Thus, the solution to the equation is \( x = 16 \).
Solve: \( \sqrt{x} - \sqrt[4]{x} - 2 = 0 \)
- Answer
-
\( x = 16 \)
While Example \( \PageIndex{ 10 } \) is interesting, the following example showcases a more frequent occurence.
Solve: \( x^{-2/3} - 3x^{-1/3} - 10 = 0 \).
- Solution
-
If we let \( u = x^{-1/3} \), then \( u^2 = \left( x^{-1/3} \right)^2 = x^{-2/3} \) and we can write\[ \begin{array}{rrclcl}
& x^{-2/3} - 3x^{-1/3} - 10 & = & 0 & & \\[6pt]
\implies & u^2 - 3u - 10 & = & 0 & \quad & \left( \text{substituting} \right) \\[6pt]
\implies & (u - 5)(u + 2) & = & 0 & \quad & \left( \text{factoring the trinomial} \right) \\[6pt]
\implies & \left(x^{-1/3} - 5\right)\left(x^{-1/3} + 2\right) & = & 0 & \quad & \left( \text{resubstituting} \right) \\[6pt]
\end{array} \nonumber \]Therefore, \( x^{-1/3} = 5 \) or \( x^{-1/3} = -2 \).\[ \begin{array}{rrcl|rclcl}
& x^{-1/3} & = & 5 & x^{-1/3} & = & -2 & & \\[6pt]
\implies & \dfrac{1}{x^{1/3}} & = & 5 & \dfrac{1}{x^{1/3}} & = & -2 & \quad & \left( \text{Laws of Exponents} \right)\\[6pt]
\implies & x^{1/3} & = & \dfrac{1}{5} & x^{1/3} & = & -\dfrac{1}{2} & \quad & \left( \text{if }a=b,\text{ where neitehr are zero, then }\frac{1}{a} = \frac{1}{b} \right) \\[6pt]
\implies & \left(x^{1/3}\right)^3 & = & \left(\dfrac{1}{5}\right)^3 & \left(x^{1/3}\right)^3 & = & \left(-\dfrac{1}{2}\right)^3 & \quad & \left( \text{raising both sides to the third power} \right) \\[6pt]
\implies & x^{3/3} & = & \dfrac{1^3}{5^3} & x^{3/3} & = & -\dfrac{1^3}{2^3} & \quad & \left( \text{Laws of Exponents} \right) \\[6pt]
\implies & x & = & \dfrac{1}{125} & x & = & -\dfrac{1}{8} & \quad & \left( \text{simplifying} \right) \\[6pt]
\end{array} \nonumber \]The domain of the original equation only has the restriction that \( x \neq 0 \) (can you see why?). Furthermore, the solution process involved cubing, not squaring. Squaring can introduce extraneous solutions, but cubing does not, so a check is not necessary. We leave it to the reader to check that these solutions work in the original equation.
Applications Involving Radicals
As you progress through your college courses, you'll encounter formulas that include radicals in many disciplines. The following Problem Solving Strategy is incredibly helpful as you progress through College Algebra and into Calculus.
- Read the problem carefully and get clarification on words or concepts that are foreign to you.
- Understand what is being stated and asked by rephrasing statements into language you connect with. Draw a figure to represent the situation, if possible.
- Label the given and wanted information. The wanted information should have a variable representing it.
- Translate into an equation relating the given and wanted information by writing the appropriate formula or model for the situation. Substitute in the given information.
- Solve the equation using good Algebra techniques.
- Check the answer in the problem and make sure it makes sense.
- State the answer to the question with a complete sentence.
One application of radicals has to do with the effect of gravity on falling objects. The following (non-memorizable) formula allows us to determine how long it will take a fallen object to hit the ground.
On Earth, if an object is dropped from a height of \(h\) feet (where \( h \) is not too large) and if we ignore air resistance, the time (in seconds) it will take to reach the ground is approximated by using the formula\[ t=\dfrac{\sqrt{h}}{4}.\nonumber \]
For example, if an object is dropped from a height of \(64\) feet, we can find the time it takes to reach the ground by substituting \(h=64\) into the formula.\[ \begin{array}{rrclcl}
& t & = & \dfrac{\sqrt{h}}{4} & & \\[6pt]
\implies & t & = & \dfrac{\sqrt{64}}{4} & \quad & \left( \text{substituting} \right) \\[6pt]
\implies & t & = & \dfrac{8}{4} & \quad & \left( \text{simplifying} \right) \\[6pt]
\implies & t & = & 2 & \quad & \left( \text{simplifying} \right) \\[6pt]
\end{array} \nonumber \]Therefore, it would take \(2\) seconds for an object dropped from a height of \(64\) feet to reach the ground (ignoring deceleration from air resistance).
The previous theorem is (for most instructors) a non-memorizable theorem. It's just a shortcut to approximate the time it takes for an object to fall a certain distance (and, in truth, it is only partially accurate for short falls).
Kaitlyn dropped her sunglasses from a bridge \(400\) feet above a river. How many seconds does it take for the sunglasses to reach the river?
- Solution
-
- Read the problem.
- Understand the situation. We are looking for the time it takes for Kaitlyn's sunglasses to reach the river.
-
Label
the given and wanted information. Let \( t \) represent the unknown time it will take the sunglasses to fall to the river.\[ \begin{array}{c|c}
\textbf{Given} & \textbf{Want} \\[6pt]
\hline h = 400 & t \\[6pt]
\end{array} \nonumber \] - Translate into an equation. The recent theorem allows us to relate \( h \) and \( t \) by the formula\[ t = \dfrac{\sqrt{h}}{4}. \nonumber \]Substituting in the given information, we get\[ t = \dfrac{\sqrt{400}}{4}. \nonumber \]
-
Solve
the equation.\[ \begin{array}{rclcl}
t & = & \dfrac{\sqrt{400}}{4} & & \\[6pt]
& = & \dfrac{20}{4} & \quad & \left( \text{simplifying} \right) \\[6pt]
& = & 5 & \quad & \left( \text{simplifying} \right) \\[6pt]
\end{array} \nonumber \] -
Check
the solution.\[ \begin{array}{rrcl}
& 5 & \overset{\text{?}}{=} & \dfrac{\sqrt{400}}{4} \\[6pt]
\implies & 5 & \overset{\text{?}}{=} & \dfrac{20}{4} \\[6pt]
\implies & 5 & = & 5 \, \checkmark \\[6pt]
\end{array} \nonumber \]Thus, \( 5 \) is correct mathematically . It also makes sense logically (for example, if the answer was negative, that would not make sense). - State the answer. It takes \( 5 \) seconds for Kaitlyn's sunglasses to reach the river.
A window washer dropped a squeegee from a platform \(196\) feet above the sidewalk. How many seconds does it take for the squeegee to reach the sidewalk?
- Answer
-
\(3.5\) seconds
Police officers investigating car accidents measure the length of the skid marks on the pavement. Then they use square roots to determine the speed, in miles per hour, a car was going before applying the brakes.
If the length of the skid marks is \(d\) feet, then the speed, \(s\), of the car (in miles per hour) before the brakes were applied can be found by using the formula\[s=\sqrt{24 d}. \nonumber \]
Again, this formula is not memorizable and it is only accurate under special circumstances (e.g., it is not accurate at all if the roadway was covered in ice).
After a car accident, the skid marks for one car measured \(190\) feet. Find the speed of the car before the brakes were applied. Round your answer to the nearest tenth.
- Solution
-
- Read!
- Understand: We are looking for the initial speed of the car before the brakes were applied.
-
Label:
Let \( s \) represent the unknown speed of the car and \( d \) represent the length of the skid.\[ \begin{array}{c|c}
\textbf{Given} & \textbf{Want} \\[6pt]
\hline d = 190 & s \\[6pt]
\end{array} \nonumber \] - Translate: \[ s = \sqrt{24 d} = \sqrt{24(190)}. \nonumber \]
-
Solve:
\[ \begin{array}{rclcl}
s & = & \sqrt{24(190)} & & \\[6pt]
& = & \sqrt{4560} & \quad & \left( \text{simplifying} \right) \\[6pt]
& \approx & 67.5277... & \quad & \left( \text{approximating} \right) \\[6pt]
& \approx & 67.5 & \quad & \left( \text{rounding to the nearest tenth} \right) \\[6pt]
\end{array} \nonumber \] -
Check:
\[ \begin{array}{rrcl}
& 67.5 & \overset{\text{?}}{\approx} & \sqrt{24(190)} \\[6pt]
\implies & 67.5 & \overset{\text{?}}{\approx} & \sqrt{4560} \\[6pt]
\implies & 67.5 & \approx & 67.5277... \, \checkmark \\[6pt]
\end{array} \nonumber \]Thus, \( 67.5 \) is correct mathematically . It also makes sense logically that the car could have been going \( 67.5 \) miles per hour. - State: The speed of the car before the brakes were applied was roughly \(67.5\) miles per hour.
An accident investigator measured the skid marks of the car. The length of the skid marks was \(76\) feet. Find the speed of the car before the brakes were applied. Round your answer to the nearest tenth.
- Answer
-
\(42.7\) mph
Access these online resources for additional instruction and practice with solving radical equations.
- Solving an Equation Involving a Single Radical
- Solving Equations with Radicals and Rational Exponents
- Solving Radical Equations
- Solve Radical Equations
- Radical Equation Application
Key Concepts
-
Binomial Squares
\(\begin{array}{l}{(a+b)^{2}=a^{2}+2 a b+b^{2}} \\[6pt] {(a-b)^{2}=a^{2}-2 a b+b^{2}}\end{array}\) -
Solve a Radical Equation
- Isolate one of the radical terms on one side of the equation.
- Raise both sides of the equation to the power of the index.
-
Are there any more radicals?
If yes, repeat Step 1 and Step 2 again.
If no, solve the new equation. - Check the answer in the original equation.
-
Problem Solving Strategy
- Read the problem carefully and get clarification on words or concepts that are foreign to you.
- Understand what is being stated and asked by rephrasing statements into language you connect with. Draw a figure to represent the situation, if possible.
- Label the given and wanted information. The wanted information should have a variable representing it.
- Translate into an equation relating the given and wanted information by writing the appropriate formula or model for the situation. Substitute in the given information.
- Solve the equation using good Algebra techniques.
- Check the answer in the problem and make sure it makes sense.
- State the answer to the question with a complete sentence.
-
Falling Objects
- On Earth, if an object is dropped from a height of \(h\) feet, the time in seconds it will take to reach the ground is found by using the formula \(t=\frac{\sqrt{h}}{4}\).
-
Skid Marks and Speed of a Car
- If the length of the skid marks is \(d\) feet, then the speed, \(s\), of the car before the brakes were applied can be found by using the formula \(s=\sqrt{24d}\).
Glossary
- radical equation
- An equation in which a variable is in the radicand of a radical expression is called a radical equation.


3.1C: Refraction and Magnification
- Describe refraction and distinguish between convex and concave lenses
The underlying principle of a microscope is that lenses refract light which allows for magnification. Refraction occurs when light travels through an area of space that has a changing index of refraction. The simplest case of refraction occurs when there is an interface between a uniform medium with an index of refraction and another medium with an index of refraction.
Some media have an index of refraction that varies gradually with position. Therefore, light rays curve through the medium rather than traveling in straight lines. This effect is what is responsible for mirages seen on hot days where the changing index of refraction of the air causes the light rays to bend creating the appearance of specular reflections in the distance (as if on the surface of a pool of water).
Taking advantage of the principle of refraction, devices can be built that can focus light. A device that produces converging or diverging light rays due to refraction is known as a lens. In general, two types of lenses exist: convex lenses , which cause parallel light rays to converge, and concave lenses , which cause parallel light rays to diverge. The former property of convex lenses is of special interest to microbiologists. In essence, a convex lens allows magnification. Light reflecting off an object is focused to a point. The simplest example of this that most people know is a magnifying glass. A magnifying glass is one convex lens, and this by itself allows the magnification of objects.
A microscope is basically a series of lenses that take advantage of the nature of refraction. Due to the nature of light, and the maximum amount of refraction that can be possible by a material, there are limits to the amount of magnification that can be done by a light microscope.
Summary
- Convex lenses allow light to converge.
- Concave lenses spread light that travels through it.
- There are limits to the amount of refraction that can be done by a material, and therefore, limits to the amount a microscope can magnify a sample.
Key Terms
- concave : Curved like the inner surface of a sphere or bowl.
- convex : Curved or bowed outward like the outside of a bowl or sphere or circle.
- specular : Pertaining to mirrors; mirror-like, reflective.


1.1: Introduction
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- Ly-Huong T. Pham, Tejal Desai-Naik, Laurie Hammond, & Wael Abdeljabbar
- ASCCC Open Educational Resources Initiative (OERI)
Introduction
In the course of a given day, think of activities that you do to entertain yourself, deliver a work product, purchase something, or interact with your family, friends, or co-workers. How many times do you snap a picture, post a text, or email your friends? Can you even remember the number of times you used a search engine in a day? Consider what you are using to do these activities. Most likely, many, if not all, of these activities involve using technologies such as a smartphone, a laptop, a website, or an app. These activities are also enabled by Wi-Fi networks that surround us everywhere, be it on the school’s campus, workplace, the airport, or even cars. You are already a user of one or more information systems, using one or more electronic devices, different software, or apps, and connect globally through different networks. Welcome to the world of information systems!
Information systems affect our personal, career, society, and the global economy by evolving to change businesses and the way we live. To prepare yourself to participate in developing or using information, building a business, or advancing your career, you must be familiar with an information system's fundamental concepts.
Defining Information Systems
Students from diverse disciplines, including business, are often required to take a course to learn about information systems. Let’s start with the term Information System (IS) . What comes to your mind? Computers? Devices? Apps? Here are a few definitions from a few sources:
- “ Information Systems is an academic study of systems with a specific reference to information and the complementary networks of hardware and software that people and organizations use to collect, filter, process, create and also distribute data .” (Wikipedia Information Systems, 2020)
- “Information systems are combinations of hardware, software, and telecommunications networks that people build and use to collect, create, and distribute useful data, typically in organizational settings.” (Valacich et al., 2010)
- “Information systems are interrelated components working together to collect, process, store, and disseminate information to support decision making, coordination, control, analysis, and visualization in an organization.” (Laudon et al., 2012)
They sound similar, yet there is something different in each as well. In fact, these authors define the terms from these perspectives:
- What are the components that make up an information system? How do they work together?
- What is the role of IS in providing value to businesses and to individuals in solving their needs ?
Let’s examine each perspective.
References
Information Systems . (2017, June 05). Retrieved July 28, 2020, from https://en.Wikipedia.org/wiki/Information_system
Laudon, K.C. and Laudon, J. P. (2012). Management Information Systems , twelfth edition. Upper Saddle River, New Jersey: Prentice-Hall.
Valacich, J. and Schneider, C. (2010). Information Systems Today – Managing in the Digital World , fourth edition. Upper Saddle River, New Jersey: Prentice-Hall.


11.4F: Antimicrobial Peptides
- Describe the role of antimicrobial peptides in host defense
Antimicrobial peptides (also called host defense peptides) are an evolutionarily conserved component of the innate immune response and are found among all classes of life. Fundamental differences exist between prokaryotic and eukaryotic cells that may represent targets for antimicrobial peptides. These peptides are potent, broad spectrum antibiotics which demonstrate potential as novel therapeutic agents.  Unlike the majority of conventional antibiotics, it appears as though antimicrobial peptides may also have the ability to enhance immunity by functioning as immunomodulators.
Antimicrobial peptides are a unique and diverse group of molecules, which are divided into subgroups on the basis of their amino acid composition and structure. Antimicrobial peptides generally consist of between 12 and 50 amino acids. These peptides include two or more positively charged residues provided by arginine, lysine or, in acidic environments, histidine, and a large proportion (generally >50%) of hydrophobic residues. The secondary structures of these molecules follow 4 themes, including:
Various AMPs : These are various antimicrobial peptide structures.
- α-helical
- β-stranded due to the presence of 2 or more disulfide bonds
- β-hairpin or loop due to the presence of a single disulfide bond and/or cyclization of the peptide chain
- Extended
Many of these peptides are unstructured in free solution, and fold into their final configuration upon partitioning into biological membranes. It contains hydrophilic amino acid residues aligned along one side and hydrophobic amino acid residues aligned along the opposite side of a helical molecule. This amphipathicity of the antimicrobial peptides allows the partition of the membrane lipid bilayer. The ability to associate with membranes is a definitive feature of antimicrobial peptides, although membrane permeabilisation is not necessary. These peptides have a variety of antimicrobial activities ranging from membrane permeabilization to action on a range of cytoplasmic targets.
The modes of action by which antimicrobial peptides kill bacteria is varied and includes disrupting membranes, interfering with metabolism, and targeting cytoplasmic components. The initial contact between the peptide and the target organism is electrostatic, as most bacterial surfaces are anionic, or hydrophobic, such as in the antimicrobial peptide Piscidin. Their amino acid composition, amphipathicity, cationic charge, and size allow them to attach to and insert into membrane bilayers to form pores by ‘barrel-stave’, ‘carpet’ or ‘toroidal-pore’ mechanisms. Alternately, they may penetrate into the cell to bind intracellular molecules which are crucial to cell living. Intracellular binding models include inhibition of cell wall synthesis, alteration of the cytoplasmic membrane, activation of autolysin, inhibition of DNA, RNA, and protein synthesis, and inhibition of certain enzymes. However, in many cases, the exact mechanism of killing is not known. One emerging technique for the study of such mechanisms is dual polarisation interferometry. In contrast to many conventional antibiotics these peptides appear to be bacteriocidal (bacteria killing) instead of bacteriostatic (bacteria growth inhibiting). In general the antimicrobial activity of these peptides is determined by measuring the minimal inhibitory concentration (MIC), which is the lowest concentration of drug that inhibits bacterial growth.
In addition to killing bacteria directly, they have been demonstrated to have a number of immunomodulatory functions that may be involved in the clearance of infection, including the ability to:
- Alter host gene expression
- Act as chemokines and/or induce chemokine production,
- Inhibit lipopolysaccharide induced pro-inflammatory cytokine production
- Promote wound healing
- Modulate the responses of dendritic cells and cells of the adaptive immune response
Animal models indicate that host defense peptides are crucial for both prevention and clearance of infection. It appears as though many peptides initially isolated and termed as “antimicrobial peptides” have been shown to have more significant alternative functions in vivo (e.g. hepcidin).
Several methods have been used to determine the mechanisms of antimicrobial peptide activity. In particular, solid-state NMR studies have provided an atomic-level resolution explanation of membrane disruption by antimicrobial peptides.
Key Points
- Antimicrobial peptides are a unique and diverse group of molecules, which are divided into subgroups on the basis of their amino acid composition and structure.
- The modes of action by which antimicrobial peptides kill bacteria is varied and includes disrupting membranes, interfering with metabolism, and targeting cytoplasmic components.
- Antimicrobial peptides have been demonstrated to have a number of immunomodulatory functions that may be involved in the clearance of infection.
Key Terms
- antimicrobial peptide : Antimicrobial peptides (also called host defense peptides) are an evolutionarily conserved component of the innate immune response and are found among all classes of life.
- innate immune : The innate immune system, also known as non-specific immune system and first line of defense, comprises the cells and mechanisms that defend the host from infection by other organisms in a non-specific manner. This means that the cells of the innate system recognize and respond to pathogens in a generic way, but unlike the adaptive immune system, it does not confer long-lasting or protective immunity to the host.
- molecules : A molecule is an electrically neutral group of two or more atoms held together by covalent chemical bonds. Molecules are distinguished from ions by their lack of electrical charge. However, in quantum physics, organic chemistry, and biochemistry, the term molecule is often used less strictly, also being applied to polyatomic ions.


12.1: Special Order Decisions
Learning Outcomes
- Create a report outlining the data to support a special order decision
In Module 11: Relevant Revenues and Costs, we went through a special order question with Hupana Running Company. Remember, they got a call to make those awesome fuchsia high tops? Here is a reminder:
| Special Order: Fuchsia High Tops | |
| Revenue (50 pair @ $150 each) | $7,500 |
| Expenses | |
| Material: Fuschia (50 × 7 units x $3.5 per unit) | $1,225 |
| Direct Labor (50 × .8 × $30) | $1,200 |
| Variable overhead (30 hours × $3 per hour) | $150 |
| Total variable expenses | $2,575 |
| Fixed Expense | |
| Dye for material cutter | $1,200 |
| Total Expenses for Special Order | $3,775 |
| Incremental net operating income | $3,725 |
So a special order is a one-time order, that is not part of a company’s regular business operation. Effective analysis needs to be done to insure that a special order will bring additional revenue to a company and not use space or time that would have been better used for the regular product lines. As a manager, it may be your responsibility to figure out if a special order should be taken, and if so, how much should be charged for the order.
In Hupana Running Company’s case, the revenue for the special order high-tops is sufficient to cover the additional costs incurred to make them. Make sure to account for things like special equipment or materials that might increase the costs. Also, make sure that you have the time available in your production schedule to handle the extra work.
Hupana needs to pay overtime to complete this order, but still made a nice profit on this special order. There may be additional things to consider from the qualitative standpoint (remember, it isn’t always about the money!). Perhaps the exposure will bring additional sales of the regular line of shoes, or other teams may want to order shoes!
Practice Questions


9.5: Outcome- Name Selection
What you’ll learn to do: explain the importance of name selection in the success of a brand
How important is naming in the success of a brand? Very important.
Consider the function of a brand name: It identifies a product, service, or company and differentiates it from competitors. But it does much more than that. It can generate attention or make something utterly forgettable. It can evoke positive or negative feelings and emotions. It can capture the imagination or drive someone to boredom. It can make a remarkable or unremarkable first impression.
Naming can be difficult in the crowded, increasingly global marketplace in which businesses operate today. As you understand the role of naming and the systematic process for selecting a new brand name, you can help lead your organization in making wise, informed choices about this essential element of branding.
The specific things you’ll learn in this section include:
- Discuss the connection between brand and name
- Outline key steps in the naming process
Learning Activities
The learning activities for this section include the following:
- Reading: Name Selection
- Self Check: Name Selection
Contributors and Attributions
- Outcome: Name Selection. Provided by : Lumen Learning. License : CC BY: Attribution


5.3: The Royal Canadian Mounted Police Act
The Royal Canadian Mounted Police Act is comparable to British Columbia’s Police Act in that it provides a legal framework that defines the way in which disciplinary procedures will be administered. The Royal Canadian Mounted Police Act contains two provisions for investigating wrongdoing by the RCMP, the External Review Committee, and the Public Complaints Commission.
The External Review Committee reviews appeals made by RCMP officers who have been disciplined under the Royal Canadian Mounted Police Act. The Public Complaints Commission is an independent team of civilians that rule on the outcome of investigations of police complaints. Critics suggest that the investigators will favour the police officer and investigate an offence with bias.
In June 2013, the federal government passed Bill C-42, which created the Enhancing Royal Canadian Mounted Police Accountability Act that is designed to enhance accountability and transparency by:
- Creating a new Civilian Review and Complaints Commission (CRCC) for the RCMP to replace the existing Commission for Public Complaints (CPC) against the RCMP and providing it with enhanced powers
- Establishing a statutory framework for handling investigations of serious incidents involving RCMP members, which will improve the transparency and accountability of these investigations
- Modernizing the RCMP’s discipline, grievance, and human resource management processes, with a view to preventing, addressing, and correcting performance and conduct issues in a timely and fair manner, which includes enabling the commissioner to establish a specific process for the investigation and resolution of harassment when a member of the Force is a respondent
This following text is a copy of an official work that is published by the Government of Canada. The reproduction has not been produced in affiliation with, or with the endorsement, of the Government of Canada. [1]
This enactment enhances the accountability of the RCMP by reforming the Royal Canadian Mounted Police Act in two vital areas. First, it strengthens the RCMP review and complaints body and implements a framework to handle investigations of serious incidents involving members. Second, it modernizes discipline, grievance, and human resource management processes for members, with a view to preventing, addressing, and correcting performance and conduct issues in a timely and fair manner.
It establishes a new complaints commission, the Civilian Review and Complaints Commission (CRCC) for the RCMP. Most notably, it sets out the authority for the CRCC to have broad access to information in the control or possession of the RCMP, it sets out the CRCC’s investigative powers, it permits the CRCC to conduct joint complaint investigations with other police complaints bodies, and it authorizes the CRCC to undertake policy reviews of the RCMP.
It establishes a mechanism to improve the transparency and accountability of investigations of serious incidents (death or serious injury) involving members, including referring the investigations to provincial investigative bodies when possible and appointing independent civilian observers to assess the impartiality of the investigations when they are carried out by the RCMP or another police service.
It modernizes the RCMP’s human resources management regime. In particular, it authorizes the Commissioner to act with respect to staffing, performance management, disputes relating to harassment, and general human resource management.
It grants the Commissioner the authority to establish a consolidated dispute resolution framework with the flexibility to build redress processes through policies or regulations. It provides for a disciplinary process that will empower managers or other persons acting as conduct authorities to impose a wide range of conduct measures in response to misconduct and that requires conduct hearings only in cases when dismissal is being sought.
Jurisdiction of the Current Commission for Public Complaints
The Commission for Public Complaints (CPC) has jurisdiction over a complaint from a member of the public that concerns the conduct of an RCMP member while performing a policing duty or function. These duties and functions include criminal investigations, public complaint investigations, policing public events, security assignments, and intelligence operations.
A complaint must also involve:
- An RCMP member or other person appointed or employed under the authority of the Royal Canadian Mounted Police Act.
- An RCMP member or other person who, when the complaint is made, is not deceased, retired, or resigned, or has been dismissed from the Force.
- Conduct that occurred after September 30, 1988, the date the CPC became authorized to take complaints.
- Enhancing Royal Canadian Mounted Police Accountability Act, http://laws-lois.justice.gc.ca/eng/a.../FullText.html ↵


16: Appendix 1 – Simple Nomenclature
You have probably encountered symbols in high school genetics and perhaps in previous chapters assigned to you in this course. Chapter 1 points out that a position on a chromosome is a gene locus, which simply means the location at which a gene is found. At that position, the gene could be one of several variants that we call alleles. The set of alleles comprise an organism’s genotype, which might affect the organism’s phenotype through the action of the genotype.
When you are solving genetics problems you have to keep track of different alleles of one or more genes. This can become confusing, so understanding how a gene works, how alleles vary, how they are expressed, and the influence they have on phenotype is important. A consistent nomenclature system – a set of rules dictating how you name alleles – becomes immensely helpful.
Figure A1.1: Two chromosomes with the same gene but different alleles, A and a. (Image used with permission of www.genome.gov provided it includes the link http://xoax.net/mediaGallery/Allele-dat342 )
Contributors
-
Dr. Todd Nickle and Isabelle Barrette-Ng (Mount Royal University) The content on this page is licensed under CC SA 3.0 licensing guidelines.


4.3: Complex Zeros
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When finding the zeros of polynomials, at some point you’re faced with the problem \(x^{2} =-1\). While there are clearly no real numbers that are solutions to this equation, leaving things there has a certain feel of incompleteness. To address that, we will need utilize the imaginary unit, \(i\).
Definition: Imaginary number \(i\)
The most basic complex number is \(i\), defined to be \(i=\sqrt{-1}\), commonly called an imaginary number . Any real multiple of i is also an imaginary number.
Example \(\PageIndex{1}\)
Simplify \(\sqrt{-9}\).
Solution
We can separate \(\sqrt{-9}\) as \(\sqrt{9} \sqrt{-1}\). We can take the square root of 9, and write the square root of -1 as \(i\).
\[\sqrt{-9} =\sqrt{9} \sqrt{-1} =3i \nonumber\]
A complex number is the sum of a real number and an imaginary number.
Definition: Complex Numbers
A complex number is a number \(z=a+bi\), where \(a\) and \(b\) are real numbers
- \(a\) is the real part of the complex number
- \(b\) is the imaginary part of the complex number \[i=\sqrt{-1}\]
Arithmetic on Complex Numbers
Before we dive into the more complicated uses of complex numbers, let’s make sure we remember the basic arithmetic involved. To add or subtract complex numbers, we simply add the like terms, combining the real parts and combining the imaginary parts.
Example \(\PageIndex{2}\)
Add \(3-4i\) and \(2+5i\).
Solution
Adding \((3-4i)+(2+5i)\), we add the real parts and the imaginary parts
\[3+2-4i+5i\nonumber\]
\[5+i \nonumber\]
Exercise \(\PageIndex{1}\)
Subtract \(2+5i\) from \(3-4i\).
- Answer
-
\[(3-4i)-(2+5i)=1-9i \nonumber\]
We can also multiply and divide complex numbers.
Example \(\PageIndex{3}\)
Multiply: \(4(2+5i)\).
Solution
To multiply the complex number by a real number, we simply distribute as we would when multiplying polynomials.
\[\begin{align*} 4(2+5i) &=4\cdot 2+4\cdot 5i \\[4pt] &=8+20i \end{align*}\]
Example \(\PageIndex{4}\)
Divide \(\dfrac{(2+5i)}{(4-i)}\).
Solution
To divide two complex numbers, we have to devise a way to write this as a complex number with a real part and an imaginary part.
We start this process by eliminating the complex number in the denominator. To do this, we multiply the numerator and denominator by a special complex number so that the result in the denominator is a real number. The number we need to multiply by is called the complex conjugate , in which the sign of the imaginary part is changed.
Here, 4+ i is the complex conjugate of 4– i . Of course, obeying our algebraic rules, we must multiply by 4+ i on both the top and bottom.
\[\dfrac{(2+5i)}{(4-i)} \, \cdot \dfrac{(4+i)}{(4+i)} \nonumber \]
To multiply two complex numbers, we expand the product as we would with polynomials (the process commonly called FOIL – “first outer inner last”). In the numerator:
\[(2+5i)(4+i)\nonumber\] Expand
\[=8+20i+2i+5i^{2}\nonumber\] Since \(i=\sqrt{-1}\), \(i^{2} =-1\)
\[=8+20i+2i+5(-1)\nonumber\] Simplify
\[=3+22i\nonumber\]
Following the same process to multiply the denominator
\[(4-i)(4+i)\nonumber\] Expand
\[=(16-4i+4i-i^{2} )\nonumber\] Since \(i=\sqrt{-1}\), \(i^{2} =-1\)
\[=(16-(-1))\nonumber\]
\[=17\nonumber\]
Combining this we get \(\dfrac{3+22i}{17} =\dfrac{3}{17} +\dfrac{22i}{17}\)
Exercise \(\PageIndex{2}\)
Multiply \(3-4i\) and \(2+3i\).
- Answer
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\[(3-4i)(2+3i)=18+i\nonumber\]
In the last example, we used the conjugate of a complex number
Definition: Complex Conjugate
The conjugate of a complex number \(a+bi\) is the number \(a-bi\).
The notation commonly used for conjugation is a bar: \[\overline{a+bi}=a-bi\nonumber\]
Complex Zeros of Polynomials
Complex numbers allow us a way to write solutions to quadratic equations that do not have real solutions.
Example \(\PageIndex{5}\)
Find the zeros of \(f(x)=x^{2} -2x+5\).
Solution
Using the quadratic formula,
\[x=\dfrac{2\pm \sqrt{(-2)^{2} -4(1)(5)} }{2(1)} =\dfrac{2\pm \sqrt{-16} }{2} =\dfrac{2\pm 4i}{2} =1\pm 2i. \nonumber\]
Exercise \(\PageIndex{3}\)
Find the zeros of \(f(x)=2x^{2} +3x+4\).
- Answer
-
\[x=\dfrac{-3\pm \sqrt{(3)^{2} -4(2)(4)} }{2(2)} =\dfrac{-3\pm \sqrt{-23} }{4} =\dfrac{-3\pm i\sqrt{23} }{4} =\dfrac{-3}{4} \pm \dfrac{\sqrt{23} }{4} i \nonumber\]
Two things are important to note. First, the zeros \(1+2i\) and \(1-2i\) are complex conjugates. This will always be the case when we find non-real zeros to a quadratic function with real coefficients.
Second, we could write
\[f(x)=x^{2} -2x+5=\left(x-\left(1+2i\right)\right)\left(x-\left(1-2i\right)\right) \nonumber\]
if we really wanted to, so the Factor and Remainder Theorems hold.
How do we know if a general polynomial has any complex zeros? We have seen examples of polynomials with no real zeros; can there be polynomials with no zeros at all? The answer to that last question, which comes from the Fundamental Theorem of Algebra, is "No."
Theorem: Fundamental theorem of algebra
A non-constant polynomial \(f\) with real or complex coefficients will have at least one real or complex zero.
This theorem is an example of an "existence" theorem in mathematics. It guarantees the existence of at least one zero, but provides no algorithm to use for finding it.
Now suppose we have a polynomial \(f(x)\) of degree \(n\). The Fundamental Theorem of Algebra guarantees at least one zero \(z_{1}\), then the Factor Theorem guarantees that \(f\) can be factored as \(f(x)=\left(x - z_{1} \right)q_{1} (x)\), where the quotient \(q_{1} (x)\) will be of degree \(n - 1\).
If this function is non-constant, than the Fundamental Theorem of Algebra applies to it, and we can find another zero. This can be repeated \(n\) times.
Theorem: complex factorization theorem
If \(f\) is a polynomial \(f\) with real or complex coefficients with degree \(n \ge 1\), then \(f\) has exactly \(n\) real or complex zeros, counting multiplicities.
If \(z_{1} ,z_{2} ,\ldots ,z_{k}\) are the distinct zero of \(f\) with multiplicities \(m_{1} ,m_{2} ,\ldots ,m_{k}\) respectively, then
\[f(x)=a\left(x-z_{1} \right)^{m_{1} } \left(x-z_{2} \right)^{m_{2} } \cdots \left(x-z_{k} \right)^{m_{k} }\]
Example \(\PageIndex{6}\)
Find all the real and complex zeros of \(f(x)=12x^{5} -20x^{4} +19x^{3} -6x^{2} -2x+1\).
Solution
Using the Rational Roots Theorem, the possible real rational roots are
\[\left\{\pm \dfrac{1}{1} ,\pm \dfrac{1}{2} ,\pm \dfrac{1}{3} ,\pm \dfrac{1}{4} ,\pm \dfrac{1}{6} ,\pm \dfrac{1}{12} \right\}\nonumber \]
Testing \(\dfrac{1}{2}\),
Success! Because the graph bounces at this intercept, it is likely that this zero has multiplicity 2. We can try synthetic division again to test that.
The other real root appears to be \(-\dfrac{1}{3}\) or \(-\dfrac{1}{4}\). Testing \(-\dfrac{1}{3}\),
Excellent! So far, we have factored the polynomial to
\[f(x)=\left(x-\dfrac{1}{2} \right)^{2} \left(x+\dfrac{1}{3} \right)\left(12x^{2} -12x+12\right)=12\left(x-\dfrac{1}{2} \right)^{2} \left(x+\dfrac{1}{3} \right)\left(x^{2} -x+1\right)\nonumber \]
We can use the quadratic formula to find the two remaining zeros by setting \(x^{2} -x+1=0\), which are likely complex zeros.
\[x=\dfrac{1\pm \sqrt{(-1)^{2} -4(1)(1)} }{2(1)} =\dfrac{1\pm \sqrt{-3} }{2} =\dfrac{1\pm i\sqrt{3} }{2} \nonumber \]
The zeros of the function are \(x=\dfrac{1}{2} ,-\dfrac{1}{3} ,\dfrac{1+i\sqrt{3} }{2} ,\dfrac{1-i\sqrt{3} }{2}\). We could write the function fully factored as \(f(x)=12\left(x-\dfrac{1}{2} \right)^{2} \left(x+\dfrac{1}{3} \right)\left(x-\dfrac{1+i\sqrt{3} }{2} \right)\left(x-\dfrac{1-i\sqrt{3} }{2} \right)\).
When factoring a polynomial like we did at the end of the last example, we say that it is factored completely over the complex numbers , meaning it is impossible to factor the polynomial any further using complex numbers. If we wanted to factor the function over the real numbers , we would have stopped at \(f(x)=12\left(x-\dfrac{1}{2} \right)^{2} \left(x+\dfrac{1}{3} \right)\left(x^{2} -x+1\right)\). Since the zeros of \(x^{2} -x+1\) are nonreal, we call \(x^{2} -x+1\) an irreducible quadratic meaning it is impossible to break it down any further using real numbers.
It turns out that a polynomial with real number coefficients can be factored into a product of linear factors corresponding to the real zeros of the function and irreducible quadratic factors which give the nonreal zeros of the function. Consequently, any nonreal zeros will come in conjugate pairs, so if \(z\) is a zero of the polynomial, so is \(\bar{z}\).
Exercise \(\PageIndex{4}\)
Find the real and complex zeros of \(f(x)=x^{3} -4x^{2} +9x-10\).
- Answer
-
Cauchy’s Bound limits us to the interval [-11, 11]. The rational roots theorem gives a list of potential zeros: \(\left\{\pm 1,\pm 2,\pm 5,\pm 10\right\}\). A quick graph shows that the likely rational root is \(x = 2\).
Verifying this,
So \(f(x)=(x-2)(x^{2} -2x+5)\)
Using quadratic formula, we can find the complex roots from the irreducible quadratic.
\[x=\dfrac{-(-2)\pm \sqrt{(-2)^{2} -4(1)(5)} }{2(1)} =\dfrac{2\pm \sqrt{-16} }{2} =\dfrac{-2\pm 4i}{2} =-1\pm 2i\nonumber \]
The zeros of this polynomial are \(x=2,\; -1+2i,\; -1-2i\)
Important Topics of This Section
- Complex and Imaginary numbers
- Finding Complex zeros of polynomials


23.12: Carbonyl Condensations with Enamines - The Stork Reaction
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Objectives
After completing this section, you should be able to
- write an equation to illustrate the three‑step Stork enamine reaction.
- write a detailed mechanism for each of the three steps of the Stork enamine reaction.
- identify the product formed, and the various intermediates (i.e., the enamine, the Michael‑type adduct), in a given Stork enamine reaction.
- identify the reagents needed to synthesize a given compound by a Stork enamine reaction.
Key Terms
- Stork enamine reaction
Study Notes
If we try to use a monoketone as a donor molecule in a Michael reaction, we will obtain a poor yield. An alternative route to the product that would be expected from such a synthesis is via the Stork enamine reaction. You may wish to review the formation of enamines (from ketones and secondary amines) before you proceed with this section; if so, review Section 19.8.
Synthesis of Enamines
As previously seen in Section 19.8 , aldehydes and ketones react with 2 o amines to reversibly form enamines. Enamines can add to acid halides to form 1,3-diketones and α, β-unsaturated Michael acceptors to from 1,5-dicarbonyls.
Reversibility
Typically the following 2 o amines are used for enamine reactions
Enamines have resonance structures similar to enolates. The lone pair electrons on the nitrogen are conjugated with the double bond and can donate electron density to the α-carbon. This allows the α-carbon of enamines to be nucleophilic in much the same fashion as enolates. The increased electron density of the α-carbon enamine, N,N-Dimethylaminoethylene, is shown as a yellow/green color in its electrostatic map. This is analogous to the increased electron density of the α-carbon of an enolate.
Enamines act as nucleophiles in a fashion similar to enolates. Because of this enamines can be used as synthetic equivalents as enolates in many reactions. This process requires a three steps: 1) Formation of the enamine, 2) Reaction with an electrophile to form an iminium salt, 3) Hydrolysis of the iminium salt to reform the aldehyde or ketone. Some of the advantages of using an enamine over an enolate are enamines are neutral, easier to prepare, and usually prevent the overreaction problems plagued by enolates. Also, enamine are more effective at the Michael reaction compared to a mono-carbonyl enolate.
Acylation of Enamines
Acylation of Enamines involves the addition of an acyl group (\(\ce{RCO-}\)) to an enamines, resulting in the formation of an α-acylated product.
Mechanism on Enamine Acylation
Step 1 Formation of the enamine
The mechanism starts with the formation of an enamine.
Step 2: Nucleophilic attack
The enamine adds to the electrophilic carbonyl carbon of the acid halide to form an iminium bond with tetrahedral alkoxide as an intermediate.
Step 3: Leaving group removal
The alkoxide reforms the carbonyl bond and eliminate a halide anion as a leaving group.
Step 4: Reform the carbonyl by hydrolysis
The iminium bond is then hydrolyzed to reform the carbonyl to create a 1,3-dicarbonyl compound as the product of a nucleophilic acyl substitution.
Michael Reaction Using Enamines: The Stork Reaction
Enamines add to α, β-unsaturated carbonyls in a Michael-like reaction. The net reaction is the addition of a ketone to a α, β-unsaturated carbonyl to product a 1,5 dicarbonyl compound as the end product. This reaction is commonly known as the Stork enamine reaction after Gilbert Stork of Columbia University who originated the work.
Mechanism of the Stork Reaction
Step 1: Formation of the enamine
Step 2: Nucleophilic attack on the carbon β to the carbonyl
After formation, the enamine adds to the electrophilic alkene carbon of the a α, β-unsaturated carbonyl form an iminium bond. The pi electrons of the alkene are pushed onto the oxygen through conjugation to form an alkoxide as an intermediate.
Step 3: Protonation
Protonation of the alkoxide forms an enol.
Step 4: Tautomerization
The enol undergoes tautomerization to reform the ketone.
Step 4: Reform the carbonyl by hydrolysis
In the last step of the mechanism, the iminium bond is hydrolyzed to reform the carbonyl to create a 1,5-dicarbonyl compound as the product of a Michael-like reaction.
Planning a Synthesis Using the Stork Enamine Synthesis
Planning a synthesis using a Stork enamine reaction is very similar to that of a Michael reaction ( Section 23-10 ). A target molecule can possibly be made using a Stork enamine reaction if it contains a 1,5-dicarbonyl. Like the Michael reaction, the key bond cleavage is a C-C bond between beta and gamma carbons from a carbonyl-like group. The fragment with the Y group loses an alpha-hydrogen and then forms a C=C bond between the alpha and beta carbon. The carbon of the other fragment gains a hydrogen. This fragment should possess an acidic alpha-hydrogen and should be made up of Michael donor fragments.
Worked Example
How could the following molecule be made using a Michael Reaction?
- Answer
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Pathway 1
Solution
Alkylation of an Enamine
Enamines undergo an S N 2 reaction with reactive alkyl halides to give the iminium salt. The iminium salt can be hydrolyzed back into the carbonyl. The individual steps are below.
Step 1: Formation of an enamine
Step 2: S N 2 Alkylation
Step 3: Reform the carbonyl by hydrolysis
All three steps together:
Exercise \(\PageIndex{1}\)
Draw the product of the reaction with the enamine prepared from cyclopentanone and pyrrolidine, and the following molecules.
- Answer
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Exercise \(\PageIndex{2}\)
What would be the starting materials necessary to make the following molecules during a Stork enamine reaction.
- Answers
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8.11: Exercise- Deciding on a Threshold for Eye Movements
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To decide on the threshold for rejecting trials with eye movements, we need to return to our goals for artifact rejection. What threshold maximizes our data quality while avoiding confounds in our data?
Let’s first consider whether horizontal eye movements are a confound in this experiment. Specifically, might horizontal eye movements cause different voltages at our FCz electrode site on deviant trials relative to standard trials? This is unlikely for two reasons. First, because FCz is on the midline, it should be near the line of zero voltage between the positive and negative sides of the voltage field produced by horizontal eye movements. Second, there is no reason to suspect that the frequency of leftward versus rightward eye movements would differ between deviants and standards. However, this is just an assumption, and we should check to make sure.
We can assess this assumption by looking at the ERP waveforms without any artifact rejection. To do this, select the dataset for Subject 10 that was created prior to any artifact detection ( 10_MMN_preprocessed_filt_be ) and then select EEGLAB > ERPLAB > Compute averaged ERPs . If you plot the resulting ERP waveforms, you’ll see a large voltage deflection for the deviants in the VEOG-bipolar channel (indicated by the green box in Screenshot 8.6). You can also see that this voltage is opposite in polarity below the eyes (VEOG-lower) versus above the eyes (Fp1 and Fp2; see the blue boxes in Screenshot 8.6). This indicates that this participant blinked more following the deviants than following the standards, just as we saw for Subject 1 (see Figure 8.2.A). However, unlike Subject 1, Subject 10 showed this pattern even during the MMN time window, so blinks could confound the MMN effects for this participant.
Now look at the HEOG-bipolar channel (indicated by the magenta box in Screenshot 8.6). The differences between deviants and standards in that channel are not any larger than the noise deflections in the prestimulus baseline period. This tells us that we don’t have to worry about differences between deviants and standards in the frequency of leftward versus rightward eye movements, confirming our assumption. This means that we mainly need to be concerned about whether the eye movements are a source of noise, not a confound.
To assess noise, we can ask how the artifact rejection impacted the SME values. Specifically, we’ll look at the SME values after rejecting only trials containing blinks, rejecting trials with blinks and eye movements using a 32 µV eye movement threshold, and rejecting trials with blinks and eye movements using a 16 µV eye movement threshold.
To start, let’s get the SME values after rejecting trials with blinks but before rejecting trials with eye movements. Make the dataset named 10_MMN_preprocessed_filt_be_ noblink s active, and EEGLAB > ERPLAB > Compute data quality metrics (without averaging) . In the Data Quality section of the GUI, select Custom parameters , click the Set DQ options… button, and create a custom time range of 125-225 ms. Make sure that the main GUI is set to exclude epochs marked during artifact detection, and then click RUN . In the data quality table that appears, look at the aSME values for Bin 1 and Bin 2 from the FCz channel in the 125-225 ms time range. These values will be our reference points for asking whether rejecting trials with eye movements makes the data quality better (because of less random variation in voltage) or worse (because of a reduction in the number of epochs being averaged together). Keep the data quality window open so that you can refer to it later.
Now repeat this process with the datasets in which eye movements were flagged for rejection using a threshold of 32 µV ( 10_MMN_preprocessed_filt_be_noblinks_HEOG32 ) and 16 µV ( 10_MMN_preprocessed_filt_be_noblinks_HEOG16 ). Compare the resulting data quality tables with the data quality table you obtained without rejection, focusing on the aSME values for Bin 1 and Bin 2 from the FCz channel in the 125-225 ms time range.
These values are summarized in Table 8.1. You can see that the data quality was slightly reduced (i.e., the aSME was increased) when large eye movements were rejected by means of the 32 µV threshold and substantially reduced when virtually all eye movements were rejected by means of the 16 µV threshold. Given that the previous analyses indicated that horizontal eye movements were not a confound, the 16 µV threshold appears to be taking us farther from the truth rather than closer to the truth (because it impairs our ability to precisely measure MMN amplitude). The 32 µV threshold decreases the data quality only slightly (probably because there is some benefit of reduced noise but some cost of a smaller number of trials). I would be inclined to go with this 32 µV threshold (instead of not excluding trials with horizontal eye movements), even though it slightly reduces the data quality, just in case there is some small confounding effect of large eye movements that wasn’t obvious.
|
Rejection |
aSME for Deviants |
aSME for Standards |
% Rejected |
|---|---|---|---|
|
Blinks Only |
0. 8264 |
0.5353 |
31.8% |
|
Blinks + Eye Movements (32 µV) |
0. 8335 |
0.5874 |
41.5% |
|
Blinks + Eye Movements (16 µV) |
0.9901 |
0.7437 |
67.6% |
Viewing a Summary of Artifacts
Table 1 shows the percentage of rejected trials. This information was printed to the Matlab Command Window when the data quality metrics were computed (and are also printed when you average). If you want to see this information for a given dataset at a later time, select the relevant dataset and then select EEGLAB > ERPLAB > Summarize artifact detection > Summarize EEG artifacts in a table . You’ll then be asked where you want to save the summary. I usually choose Show at Command Window.
As you have seen, there is some subjectivity involved with artifact rejection. In my experience, a well-trained researcher can meaningfully increase the final data quality and avoid confounds by carefully setting the artifact detection parameters individually for each participant in this manner. It takes some time, but you will get much faster as you gain experience. The two participants we’ve examined so far in this chapter are particularly challenging cases that require some careful thought and analysis, but most of the participants in this study were much more straightforward. I find that we can use a standard set of detection parameters in about 80% of participants in my lab’s basic science experiments, and it takes only 5–10 minutes to verify that everything is working fine in these participants.
Beyond the time investment, it’s also important to consider whether customizing the artifact rejection for each participant might lead to some kind of bias in the results. Most basic science ERP studies involve within-subjects manipulations, in which the same artifact detection parameters are used for all conditions for a given participant. Because the parameters are identical across conditions, there is little opportunity for bias. In theory, the experimenter could try many different artifact detection parameters for a given participant and then choose the parameters that produce the desired effect. But this is obviously cheating. If someone wants to cheat, there are much easier ways to do it, so I don’t worry much about this possibility. To avoid unconscious bias, you should avoid looking at the experimental effects in the averaged ERP waveforms when you’re setting the parameters (although you may need to look at the averaged EOG waveforms to assess the presence of systematic differences in artifacts between conditions).
My advice is different for research that focuses on comparing different groups of participants (e.g., a patient group and a control group). In these studies, the main comparisons are between participants, and now we may have different artifact detection parameters for our different groups. This could lead to unintentional biases in the results. To minimize any biases, I recommend that the person setting the artifact detection parameters for the individual participants should be blind to group membership. For example, in my lab’s research on schizophrenia, the person setting the artifact detection parameters is blind to whether a given participant is in the schizophrenia group or the control groups. That’s a bit of a pain, but it’s worth it to avoid biasing the results. Note that some subjectivity also arises in artifact correction (e.g., choosing which ICA components to eliminate), so the person doing the correction should also be blind to group membership.
When to set artifact detection parameters (and how to avoid a catastrophe)
Imagine that you spend 9 months collecting data for an ERP study, and at the end you realized that there was a major problem with the data that prevented you from answering the question that the study was designed to answer. Your heart would start racing.  You would feel like vomiting. And you would want to crawl into a hole and never come out.
Would you like to avoid that situation? If so, then here’s an important piece of advice: Do the initial processing of each participant’s data within 48 hours of the recording session. This includes every step through averaging the data and examining the averaged ERPs. Of course, this includes setting the artifact detection parameters. And it includes quantifying the behavioral effects (which is the step that people most frequently forget).
If you don’t do this, there is a very good chance that there will be a problem with your event codes, or with artifacts, or with something unique to your experiment that I can’t anticipate, and that this problem will make it impossible for you to analyze your data at the end of the study. I have seen this happen many, many, many times. Many times!
You can catch a lot of these problems by doing a thorough analysis of the first participant’s data before you run any additional participants. And in my lab, we have a firm rule that experimenters aren’t even allowed to schedule the second participant until we’ve done a full analysis of the first participant’s data. I estimate that we catch a problem about 80% of the time when we analyze the first participant’s data. So you absolutely must do this.
However, some problems don’t become apparent until the 5 th or the 15 th participant. And sometimes a new problem arises midway through the study. For this reason, you really must analyze the data from each participant within a couple days.
There is another side benefit to this: You won’t be in the position of needing to set the parameters for 30 participants in a single two-day marathon preprocessing session. Not only would these be two of the most dreariest days of your life, it would be difficult for you to pay close attention and do a good job of setting the parameters. The task of setting the parameters is best distributed over time.


6.1: Linear Transformations
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Outcomes
- Understand the definition of a linear transformation, and that all linear transformations are determined by matrix multiplication.
Recall that when we multiply an \(m\times n\) matrix by an \(n\times 1\) column vector, the result is an \(m\times 1\) column vector. In this section we will discuss how, through matrix multiplication, an \(m \times n\) matrix transforms an \(n\times 1\) column vector into an \(m \times 1\) column vector.
Recall that the \(n \times 1\) vector given by \[\vec{x} = \left [ \begin{array}{r} x_1 \\ x_2\\ \vdots \\ x_n \end{array} \right ]\nonumber \] is said to belong to \(\mathbb{R}^n\), which is the set of all \(n \times 1\) vectors. In this section, we will discuss transformations of vectors in \(\mathbb{R}^n.\)
Consider the following example.
Example \(\PageIndex{1}\): A Function Which Transforms Vectors
Consider the matrix \(A = \left [ \begin{array}{ccc} 1 & 2 & 0 \\ 2 & 1 & 0 \end{array} \right ] .\) Show that by matrix multiplication \(A\) transforms vectors in \(\mathbb{R}^3\) into vectors in \(\mathbb{R}^2\).
Solution
First, recall that vectors in \(\mathbb{R}^3\) are vectors of size \(3 \times 1\), while vectors in \(\mathbb{R}^{2}\) are of size \(2 \times 1\). If we multiply \(A\), which is a \(2 \times 3\) matrix, by a \(3 \times 1\) vector, the result will be a \(2 \times 1\) vector. This what we mean when we say that \(A\) transforms vectors.
Now, for \(\left [ \begin{array}{c} x \\ y \\ z \end{array} \right ]\) in \(\mathbb{R}^3\), multiply on the left by the given matrix to obtain the new vector. This product looks like \[\left [ \begin{array}{rrr} 1 & 2 & 0 \\ 2 & 1 & 0 \end{array} \right ] \left [ \begin{array}{r} x \\ y \\ z \end{array} \right ] = \left [ \begin{array}{c} x+2y \\ 2x+y \end{array} \right ]\nonumber \] The resulting product is a \(2 \times 1\) vector which is determined by the choice of \(x\) and \(y\). Here are some numerical examples. \[\left [ \begin{array}{ccc} 1 & 2 & 0 \\ 2 & 1 & 0 \end{array} \right ] \left [ \begin{array}{c} 1 \\ 2 \\ 3 \end{array} \right ] = \ \left [ \begin{array}{c} 5 \\ 4 \end{array} \right ]\nonumber \] Here, the vector \(\left [ \begin{array}{c} 1 \\ 2 \\ 3 \end{array} \right ]\) in \(\mathbb{R}^3\) was transformed by the matrix into the vector \(\left [ \begin{array}{c} 5 \\ 4 \end{array}\right ]\) in \(\mathbb{R}^2\).
Here is another example: \[\left [ \begin{array}{rrr} 1 & 2 & 0 \\ 2 & 1 & 0 \end{array} \right ] \left [ \begin{array}{r} 10 \\ 5 \\ -3 \end{array} \right ] = \ \left [ \begin{array}{r} 20 \\ 25 \end{array} \right ]\nonumber \]
The idea is to define a function which takes vectors in \(\mathbb{R}^{3}\) and delivers new vectors in \(\mathbb{R}^{2}.\) In this case, that function is multiplication by the matrix \(A\).
Let \(T\) denote such a function. The notation \(T:\mathbb{R}^{n}\mapsto \mathbb{R}^{m}\) means that the function \(T\) transforms vectors in \(\mathbb{R}^{n}\) into vectors in \(\mathbb{R}^{m}\). The notation \(T(\vec{x})\) means the transformation \(T\) applied to the vector \(\vec{x}\). The above example demonstrated a transformation achieved by matrix multiplication. In this case, we often write \[T_{A}\left( \vec{x}\right) =A \vec{x}\nonumber \] Therefore, \(T_{A}\) is the transformation determined by the matrix \(A\). In this case we say that \(T\) is a matrix transformation.
Recall the property of matrix multiplication that states that for \(k\) and \(p\) scalars, \[A\left( kB+pC\right) =kAB+pAC\nonumber \] In particular, for \(A\) an \(m\times n\) matrix and \(B\) and \(C,\) \(n\times 1\) vectors in \(\mathbb{R}^{n}\), this formula holds.
In other words, this means that matrix multiplication gives an example of a linear transformation, which we will now define.
Definition \(\PageIndex{1}\): Linear Transformation
Let \(T:\mathbb{R}^{n}\mapsto \mathbb{R}^{m}\) be a function, where for each \(\vec{x} \in \mathbb{R}^{n},T\left(\vec{x}\right)\in \mathbb{R}^{m}.\) Then \(T\) is a linear transformation if whenever \(k ,p\) are scalars and \(\vec{x}_1\) and \(\vec{x}_2\) are vectors in \(\mathbb{R}^{n}\) \(( n\times 1\) vectors\(),\) \[T\left( k \vec{x}_1 + p \vec{x}_2 \right) = kT\left(\vec{x}_1\right)+ pT\left(\vec{x}_{2} \right)\nonumber \]
Consider the following example.
Example \(\PageIndex{2}\): Linear Transformation
Let \(T\) be a transformation defined by \(T:\mathbb{R}^3\to\mathbb{R}^2\) is defined by \[T\left [\begin{array}{c} x \\ y \\ z \end{array}\right ] = \left [\begin{array}{c} x+y \\ x-z \end{array}\right ] \mbox{ for all } \left [\begin{array}{c} x \\ y \\ z \end{array}\right ] \in\mathbb{R}^3\nonumber \] Show that \(T\) is a linear transformation.
Solution
By Definition \(\PageIndex{1}\) we need to show that \(T\left( k \vec{x}_1 + p \vec{x}_2 \right) = kT\left(\vec{x}_1\right)+ pT\left(\vec{x}_{2} \right)\) for all scalars \(k,p\) and vectors \(\vec{x}_1, \vec{x}_2\). Let \[\vec{x}_1 = \left [\begin{array}{c} x_1 \\ y_1 \\ z_1 \end{array}\right ], \vec{x}_2 = \left [\begin{array}{c} x_2 \\ y_2 \\ z_2 \end{array}\right ]\nonumber \] Then \[\begin{aligned} T\left( k \vec{x}_1 + p \vec{x}_2 \right) &= T \left( k \left [\begin{array}{c} x_1 \\ y_1 \\ z_1 \end{array}\right ] + p \left [\begin{array}{c} x_2 \\ y_2 \\ z_2 \end{array}\right ] \right) \\ &= T \left( \left [\begin{array}{c} kx_1 \\ ky_1 \\ kz_1 \end{array}\right ] + \left [\begin{array}{c} px_2 \\ py_2 \\ pz_2 \end{array}\right ] \right) \\ &= T \left( \left [\begin{array}{c} kx_1 + px_2 \\ ky_1 + py_2 \\ kz_1 + pz_2 \end{array}\right ] \right) \\ &= \left [\begin{array}{c} (kx_1 + px_2) + (ky_1 + py_2) \\ (kx_1 + px_2)- (kz_1 + pz_2) \end{array}\right ] \\ &= \left [\begin{array}{c} (kx_1 + ky_1) + (px_2 + py_2) \\ (kx_1 - kz_1) + (px_2 - pz_2) \end{array}\right ] \\ &= \left [\begin{array}{c} kx_1 + ky_1 \\ kx_1 - kz_1 \end{array}\right ] + \left [ \begin{array}{c} px_2 + py_2 \\ px_2 - pz_2 \end{array}\right ] \\ &= k \left [\begin{array}{c} x_1 + y_1 \\ x_1 - z_1 \end{array}\right ] + p \left [ \begin{array}{c} x_2 + y_2 \\ x_2 - z_2 \end{array}\right ] \\ &= k T(\vec{x}_1) + p T(\vec{x}_2) \end{aligned}\nonumber \] Therefore \(T\) is a linear transformation.
Two important examples of linear transformations are the zero transformation and identity transformation. The zero transformation defined by \(T\left( \vec{x} \right) = \vec{0}\) for all \(\vec{x}\) is an example of a linear transformation. Similarly the identity transformation defined by \(T\left( \vec{x} \right) = \vec{x}\) is also linear. Take the time to prove these using the method demonstrated in Example \(\PageIndex{2}\) .
We began this section by discussing matrix transformations, where multiplication by a matrix transforms vectors. These matrix transformations are in fact linear transformations.
It turns out that every linear transformation can be expressed as a matrix transformation, and thus linear transformations are exactly the same as matrix transformations.