Lesson: Absolute Value Equations

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Comment on Absolute Value Equations

Hi, Just a quick question. When you say to plug our possible solutions to the original equation, is the
-(negative) equation not considered part of the original equation? I ask because i'm confused. If{x}=3x-4 or -3x-4;
my first solution of 2 i plugged into 3x-4. my second solution of 1 i plugged into -3x-4 because I thought the solution of 1 was derived from that "original equation of -3x-4. So for future reference, when I am tackling absolute value questions, I am always plugging the solutions into the first equation only?
gmat-admin's picture

You must plug the solutions into the ORIGINAL equation (not the derived equation). Otherwise, you are just confirming what you already solved.

That is, from the equation |x| = 3x - 4, you derived two equations: x = 3x - 4, and x = -(3x - 4)

When you solved x = 3x - 4, and got x = 2, you got that solution by solving the DERIVED equation (x = 3x - 4), so it doesn't help to then plug that solution into the (derived) equation that you just used to determine that solution.

If you follow that strategy, you will never find any extraneous roots.

Our goal is to solve the original equation, so we must use that equation to check our possible solutions.

If y + |y| = 0 which of the following must be true?

A. y > 0
B. y >= 0
C. y < 0
D. y <= 0
E. y = 0
Can we chose y as + and - ? I chose y as 1 and | y | as -1 so I add 1 + (-1) = 0 . I don't understand why the answer is d .
gmat-admin's picture

"I chose y as 1 and |y| as -1"

If y = 1, then |y| = |1| = 1 (not -1).

NOTE: The absolute value of ANY value is always greater than or equal to 0. In other words, the absolute value of a number CANNOT be negative.

Let's solve the question by testing values.

Try y = 2. Plug into equation to get: 2 + |2| = 0
Evaluate: 4 = 0
So, y cannot equal 2.

Try y = 0. Plug into equation to get: 0 + |0| = 0
Evaluate: 0 = 0
So, y can equal 0.

Try y = -1. Plug into equation to get: -1 + |-1| = 0
Evaluate: 0 = 0

By the process of elimination, the correct answer is D

Hi Brent,
You used what you called U-Substitution which is great idea to simplify the original equation, but for someone like me and many non-math experts, this idea will not always come in mind.
So I tried to simplify the equation as follows:
|x-4| * |x-4| + |x-4| = 30
x^2 -4x -4x +16 + x-4 = 30
x^ -7x -18 =0
(x- 9) (x+2) = 0
x=9 or x=-2

Just wondering if the way I am using to simplify the equation that includes absolute value is valid approach?

gmat-admin's picture

Question link: https://gmatclub.com/forum/what-is-the-sum-of-all-the-real-values-of-x-f...

Hi Aladdin,

Unfortunately, that approach will tend to miss potential solutions (as it did here).

Also, as with all equations involving absolute value, you must always confirm whether any of your solutions are extraneous roots.

When we test your solution x = -2, we find that it is NOT a solution to the original equation.


Hi Brent,
I have used the method of plugging in the answer choices and I got 8 is the only number that was valid. I found it faster than the algebraic method in this case.
I am not sure if that was a good approach in this kind of questions or it was coincident.
gmat-admin's picture

Question link: https://gmatclub.com/forum/if-x-4-what-is-the-range-of-the-solutions-of-...

Hi Aladdin,

Your solution is a fortunate mistake. Yes, it yielded the correct answer, but that was just a coincidence.

I say this because you answered a question that's different from the question that is asked. The question asks us to find the RANGE of all possible solutions.

When we solve the given equation, we find out that there are 3 solutions: x = 8, x = 10 and x = 16

So, the RANGE = biggest value - smallest value
= 16 - 8
= 8

So, when you plugged in the answer choices, you found that x = 8 is ONE solution. This alone is not enough to determine the RANGE of all possible solutions.

However, since you misread the question as "What is the solution of the equation |14–x| = 24/(x−4)?", you entered answer choice C (which turns out to be the correct answer .... to the real question).

Does that help?


Thanks Brent.

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