In this video we will take a look at an advanced algebraic concept of absolute value equations. Check out this video, try the problems presented, and get a healthy handle on this concept. Then take a look at our other advanced algebra videos!

# Video Transcript: Absolute Value Equations

Now we’ll begin a section on advanced algebra, kind of a grab bag of advanced topics in algebra. The first thing we’ll talk about are absolute value equations. Recall what we said about absolute value in the lesson Positive and Negative Numbers II, in the Arithmetic and Fractions module. If you haven’t seen that module yet and you haven’t seen that lesson, I’d suggest looking at that first and getting a good sense about absolute value before we go through all the information in this lesson.

So, we’ll begin with a simple observation like for example absolute value of positive 5 of course is positive 5, absolute value of negative 5 is of course positive 5. So either of these equations, absolute value of 5 = x or absolute value of -5 = x, would have the single solution x = 5. Those are not very interesting equations, not very interesting when there are no variables inside the absolute value.

### Insert Variables

It gets more interesting when we put variables inside the absolute value. So the absolute value of x = 5 think about that, that has two solutions, positive 5 or negative 5, x could be either one of these values. And this is similar to the way the quadratic equation has two solutions. So, this is another category of algebra equations that can have two solutions. Suppose an expression other than x is inside the absolute value.

So, for example, the absolute value of the quantity 2x + 3, and all that’s gonna equal 5. Well of course, if that whole expression has an absolute value of 5, then there’s only two possibilities for it. Either, the expression inside equals 5 or equals -5, so very much the same approach.

We’re gonna set that expression equal to 5 and set it equal to -5 and, of course, this connected by the word OR. And remember exactly the same way with quadratics, this word OR is a mathematical object. It’s not a piece of garnish, it’s a necessary part of the problem. So we solve the equations individually and we get x = 1 or x = -4, and those are the two possibilities.

Generalizing, we could say that if E is any algebraic expression, and k is any positive number, then the equation absolute value of E = k has the solutions E = k or E=- k. I should hasten to add here, this notation I am using E for an expression, I just made this up on the this page, this is not a general notation that you’re gonna see anywhere else.

I’m just using a fore short hand here, the idea is that we could put any expression inside the absent values. And then whatever it equals whatever that absolute value equals the expression itself will be equal to positive or negative of that same value. Notice if the absolute value is not by itself on one side, we will have to isolate it before we can break it into the two OR equations.

## Practice Problem One

Here’s a practice problem, pause the video and then we’ll talk about this.

So of course with this we have to isolate the absolute value first we subtract 1. Once we subtract 1 then we’re all set. I’m gonna break it into two equations. Solve each equation individually divided by 3 and we get x = +2/3 or x = -2 and those are the two solutions.

What if the absolute value of an expression equals not in single number, but an other expression? So now we’re actually getting into the territory that the test would actually ask us. We’ll still follow the same plan, so absolute value of anything equals something else.

It means that if that thing inside the absolute value A equals either positive B or negative B. So it equals either the expression on the other side or the expression on the other side of the negative side in front of it. Sometimes this will give us extraneous solutions. Solutions that result correctly from the math, but which don’t work in the original equation, we have to check each solution.

So this is an odd thing about absolute value equations. And we’ll see this again much later on when we talk about square roots also–that we get extraneous solutions. That means we have to check our work, check the answers that we find.

## Practice Problem Two

So here’s a practice question. Pause the video, work on this and then we’ll talk about this.

Okay, for this one, the absolute value is already isolated, so we can immediately break this into the or equation. So we say that the thing inside the absolute value of 1 + 2x either equals that expression, 4- x, or equals the expression with a negative sign in front of it. And of course, putting a negative sign in front of subtraction just reverses the orders of the subtraction, it becomes x- 4.

So first thing we’re gonna do is get all the exes on one side in each equation, then subtract the one and isolate the x, and then divide. And we get the solutions, x = 1 and x = -5. Now we’re not done here, we did the correct math. These two solutions correctly result from the math. But we don’t know, do they actually satisfy the original equation, so we have to check.

So for x = 1, I plug it in to the left side I get 3, I plug it into the right side I get 3. That works, for -5 I plug it into the left side, I get 9, I plug it in to the right side I get 9. So that also works, both answers work here. So here we have two valid solutions of the equation.

## Practice Question Three

Here’s a practice question. Pause the video, work on this and then we’ll talk about this.

Okay, again, the absolute value is isolated so we can go directly to the or equation. So the thing inside x + 4 that is either going to equal the expression or equal the expression multiplied by a negative sign. So the first thing we’re gonna do is get all the x’s on one side, then we’re gonna subtract to isolate the x.

And we get either x equals 1 or x equals negative 1.5 or negative three-halves. So, now we’ll plug these in and check these. Those are our two solutions. For positive 1, when we plug that into the left side, we get 5. When plug it in the right side we get 5, and that checks that’s perfectly fine.

-1.5 when we plug this into the left side, we get 2.5. When we plug it in the right side, we get negative -2.5, so this doesn’t work. It gives a different value on the right and left side when we plug in. So this doesn’t work and this equation has only one solution, x = 1.

## Practice Question Four

Here’s another practice question, pause the video and then we’ll talk about this.

So we can go directly to the or equation. The thing inside, 2x + 5 either equals x + 1, or equals x + 1 multiplied by a negative sign. We’ll group the x’s on one side, we’ll subtract the 5 and we have to divide by 3 in the second equation. So we get x equals negative 4, or x equals negative 2, so these are our roots.

### Checking

So we have to check these roots now. Check the first one, x = -4, when we plug it into x + 1 we get -3. That’s a negative number. We don’t even have to plug in to the absolute value because we know the absolute value can’t have an output that’s negative. So this answer doesn’t work.

Now we will check -2 when we plug that into the right side we also get a negative number. So that also doesn’t work for the same reason. We can’t have the output of the absolute value equaling a negative number. So both of these answers resulted from correct math. We followed all of the steps of the math correctly, but both of these are extraneous roots.

The algebra leads us to think that they might be solutions. But it turns out that they don’t actually work when we plug them into the original equation. So both of our roots here are extraneous roots and what that means is the equation at the top has no solution.

In summary, we solve the equation value of A = B by splitting it into two equations, A = B or A = -B.

Remember, we have to isolate the absolute value before we can split it into the two or equations. Once we solve for answers we have to check that each works in the original equation; in this way we eliminate extraneous roots.

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