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Video Transcript (0:07) Absolute value is a concept that most students already feel that they have a good understanding of, but many cannot answer the simple question “what is absolute value?” correctly. The most common answer that students give is that absolute value makes things positive, but that is only what it does. The concept of absolute value is the concept of distance from 0. (0:31) For example, |5| = 5, since 5 is 5 units away from 0. However, |-5| also equals 5, since that is how far away it is from 0. (0:45) This distinction may not seem important to many students initially, but the conceptual understanding that this fosters extends into your understanding of complex numbers, absolute value equations, and even polar coordinates when you reach Precalculus. (1:00) The basic idea of absolute value is simple but can become immediately more useful. (1:05) For example, consider the absolute value equation |x| = 3. Verbalizing this equation with the idea of distance from 0, the equation is read as “the distance of x from 0 is 3.” This should instantly make you think of the numbers 3 and -3, since they are both the same distance from 0! (1:29) Let’s take the same example and change it slightly. What if we were given the absolute value equation |x – 1| = 3? This doesn’t seem to fit in the interpretation of absolute value being a distance that we had earlier, so maybe it wasn’t actually useful to think like that… (1:48) However, this equation can be interpreted as the distance of x from 1 is 3. The only difference from our previous example is that we have now shifted our whole solution from being centered at 0 to being centered at 1! This is equivalent to shifting the whole solution right by 1 unit, but don’t worry too much about that in this lesson. (2:10) In this example, absolute value is still distance, but now we’ve turned it into the distance from another number. (2:17) Using the same example once more, let’s try and solve this equation analytically, without relying on the number line graphics used so far. (2:25) If I have the equation |x – 1| = 3, I know that one of two possibilities exist: x – 1 = 3 x – 1 = -3 (2:39) This is due to the fact that absolute value equations will guarantee that the “chunk” inside the absolute value is either the positive or negative value of the other side of the equation. Solving each of these equations separately, we add 1 to each side and find two values for x. x = 4 x = -2 (3:01) This corresponds directly to the number line we discovered earlier! (3:08) Solving equations algebraically is very useful when the absolute value equations get much busier. For example, the equation 3|2x + 6| - 3 = 12 is not easily solvable with a number line. (3:24) However, simply executing basic arithmetic steps will result in a simpler solution. (3:30) First, add 3 to both sides to find the equation 3|2x + 6| = 15. (3:37) Divide both sides of the equation by 3 to find |2x + 6| = 5. (3:45) This is where we split into 2 cases just like earlier! (3:48) Case 1 is when 2x + 6 = 5. Subtract 6 and divide by 2 to find that x = -½. (3:57) Case 2 is when 2x + 6 = -5. Subtract 6 and divide by 2 to find that x = -11/2. (4:07) The main idea is that we must always isolate the absolute value piece of the equation to find two answers for x. (4:15) Using both a graphical understanding and developing the skills to solve these equations will allow you to tackle any absolute value problem with ease.
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Overview Absolute value, represented as a vertical line on either side of the value, refers to a number or variable’s distance from zero on the number line. There are always two numbers that share the same distance from zero: a positive number and its negative counterpart. Thus, the absolute value of a number is that number without a positive or negative sign. For example: |4| = 4 |-4| = 4 Both of these values (4 and -4) are equidistant from 0, and thus have the same absolute value. Absolute Value Examples The absolute value for a variable will usually result in a positive and a negative value as well, but the numbers will not necessarily be counterparts of each other. Instead, the possible values of the variable generate two outcomes where the equation is true. Practice Problem: Solve for 𝑥 Solution First, solve the equation but without the absolute value bars, which gives 7 + 𝑥 = 14; subtract 7 from both sides and you get 𝑥 = 7 Second, multiply the right hand side of the equation by -1 and solve again, which gives 7 + 𝑥 = -14; subtract 7 from both sides and you get 𝑥 = -21 𝑥 = 7, -21 Some absolute value problems can be quite tricky, but you can solve all absolute value questions by making sure to account for the both the positive and negative counterparts. For example, absolute value inequalities can be a bit complicated. The inequality |𝑥| < 2 represents the values of 𝑥 such that distance between 𝑥 and 0 is no more than 2, which can be written as -2 < 𝑥 < 2. On the other hand, the inequality |𝑥| > 2 represents the values of 𝑥 such that the distance between 𝑥 and 0 is at least 2, which can be written as 𝑥 < -2 and 𝑥 > 2 Practice Problem: Solve for 𝑥 Solution |3𝑥 + 9| < 18 -18 < 3𝑥 + 9 < 18 -27 < 3𝑥 < 9 -9 < 𝑥 < 3 This is a more complicated inequality; to solve it, first isolate the absolute value expression. Next, since the absolute value expression is less than 18, we know that the distance between (3𝑥 + 9) and 0 must be less than 18. Now we know that 𝑥 must be between -9 and 3. Absolute Value Practice Problems 1. Solve the following absolute value problems: |255| = ? |-9999| = ? |34 - 57| = ? 2. Solve for 𝑥 in the following absolute value problems: |4𝑥 + 2| = 12 |-𝑥 + 5| = 20 4|𝑥 - 3| = 28 3. Which of the following is the least value that solves for 𝑥 in the equation |𝑥 - 5| = 4? -9 -1 1 9 4. What are the values for 𝑥 in the equation |4𝑥 + 2| + 6 = 24? -2.5, 4.5 4, -5 2.5, -5.5 -4, 5 5. What are the values for 𝑥 in the equation |3 - 5𝑥| = 20? -17/5, 23/5 5/17, -5/23 17/5, -23/5 -5/17, 5/23 6. What are the values for 𝑥 in the equation |4𝑥| + 5 < 85? -20 < 𝑥 < 20 -20 > 𝑥 > 20 -45 < 𝑥 < 20 -45 > 𝑥 > 20 Answers 255, 9999, 23 2.5 and -3.5, -15 and 25, 10 and -4 1 4, -5 -17/5, 23/5 -20 < 𝑥 < 20
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