# Question: Absolute Value Question

## Comment on Absolute Value Question

### I have a problem the fact

I have a problem the fact that Statement 1 is insufficient. If the value of x is either 4 or -3, plugging back`those values show that only 4 could be the correct value. As such the statement is sufficient asit shows a definite value for x ### Be careful. You are using

Be careful. You are using information from statement 2 to help you in determining whether statement 1 is sufficient. This is a common mistake, which we cover in this video: https://www.gmatprepnow.com/module/gmat-data-sufficiency/video/1097

If statement 1 is sufficient, then it must be the case that the information in statement 1, ALONE, is enough to answer the target question (what is the value of x?).

If we examine statement 1 ALONE, we see that there are two possible values of x (x = 16 and x = 4), so statement 1 is NOT SUFFICIENT.
You have taken information from statement 2 (x = 4 or x = -3) and you have plugged those values into statement 1.

You must deal with each statement on an INDIVIDUAL basis (unless it turns out that both statements are insufficient on their own, in which case we examine the two statements combined).

### Can we write x^2 - 16 = x-4

Can we write x^2 - 16 = x-4 as => (x+4)(x-4) = x-4 and therefore dividing both sides by x-4? ### Unfortunately, that technique

Unfortunately, that technique won't work. When you divide both sides by (x-4), you may be inadvertently dividing both sides by 0.

When you used that technique, you would have found only one possible solution, when there are actually two possible solutions (x = 4 and x = -5)

### I am still not convinced why

I am still not convinced why we should not be dividing both the sides by (x-4). Because, that was the first thing that struck my mind while solving the equation. If you could help me understand the other way. ### I'll show you by example:

I'll show you by example:

If x = 4, then we can conclude with certainty that 7(x - 4) = 11(x - 4).

We can PROVE that this equation holds true by replacing x with 4 to get: 7(4 - 4) = 11(4 - 4).

Simplify: 7(0) = 11(0)
So, 0 - 0
Perfect.

Now let's take 7(x - 4) = 11(x - 4) and divide both sides by (x-4).

We get: 7 = 11

Hmmm, what happened?

What happened is that we took an equation and divided both sides by 0 (since x-4 is equal to zero when x = 4)

So, before we can divide ANY equation by a variable (like x-4), then we must be absolutely certain that we aren't inadvertently dividing by zero.

### This helps. I guess I had

This helps. I guess I had some wrong calculation blocking my mind. Thanks !

### you can devide by x-4 but

you can devide by x-4 but there is a risk to miss that x-4 = 0 gives also one of the solution, indeed the solution as the 2 other solutions for x are extraneous.

by simplifying by (x-4), we have effectively three equation
x - 4 = 0 => x = 4
x + 4 = -1 => x = -5 (extraneous)
x + 4 = 1 => x = -3 (extraneous)

### Brent,

Brent,

Exercise 57 (Problem Solving - OG 2017)
Could you explain me in a different way from OG 2017 answer solution?

Cheers,
Pedro ### Here's my full solution

Here's my full solution (which is different from that in the OG2017): https://gmatclub.com/forum/if-y-is-an-integer-then-the-least-possible-va...

### Brent,

Brent,

From exercise 398 (Data Sufficiency - OG 2017).

Number 1) for me it is well explained.
But number 2) isn't it insufficient? Because k^2=2.601x10^9, then k=51x10^3, but if I put k=-51x10^3, k^2 will be 2.601x10^9, right? So, I would have 2 results for k.

Please could you explain my doubt.

Cheers,
Pedro ### Be careful, √(10^9) does not

Be careful, √(10^9) does not equal 10^3
Notice that (10^3)(10^3) = 10^6
So, √(10^6) = 10^3, but √(10^9) equals something else.

Here's my step-by-step solution to this question: https://gmatclub.com/forum/if-n-is-a-positive-integer-and-k-5-1-10-n-wha...

Cheers,
Brent

### hi Brent,

hi Brent,

interesting question! I was wondering if there are any laws that can help us do the following:

x^2 - 16 = x-4
(x+4)(x-4) = x-4

from there, there are two possible cases:
either

case 1: -x+4 = x-4 or

case 2: x-4 = x-4

when we solve these

case 1: x = 4

case 2: everything cancels out - so no solutions.

bearing in mind that 4 indeed is the right answer, is this a feasible approach? is there any mathematical law that supports this thinking - thanks again for your help! ### Can you explain how you

Can you explain how you derived cases 1 and 2?

Also, your equation for case 2 (x - 4 = x - 4) actually has infinitely many solutions (you are saying that it has zero solutions).

In fact, ANY value of x will satisfy the equation x - 4 = x - 4.

For example, if we plug in x = 9, we get; 9 - 4 = 9 - 4. This works, so x = 9 is a possible solution.

Cheers,
Brent

### Hi! Could we say, since

Hi! Could we say, since statement 2 leads to a system of two equations, that only the solution shared in both eq. Will be valid, so directly disregard solutions -5 & -3 as valid? Thanks ! ### Sorry, I'm not entirely sure

Sorry, I'm not entirely sure I understand your question. That said, here's my response to what I believe you are asking:

Once we solve our resulting equations, we must be sure always test the solutions for extraneous roots.

So, even though both equations share x = 4 as a possible solution, it's also possible that x = -5 or x = -3 are solutions too. So, we must still text them to see whether they are extraneous.

Does that help?

Cheers,
Brent

Ok, thanks!

### Hi Brent,

Hi Brent,
Very thorough method for such questions! However, this is time-consuming and one could get confused with the number of solutions and equations while solving and testing by plugging the values. Is there any alternative method? ### Hi krati,

Hi krati,

Yes, it is very time consuming to test for extraneous roots.
There's really no way to avoid this.
If you're running short on time, this is a good question to skip so you can devote that time to other questions.

Cheers,
Brent

### hello sir

hello sir
Can we solve this question like this

|x-10| = 6
case 1: x - 10 ≥ 0, so x ≥ 10
x - 10 = 6
x = 16 accepted (since x has to be more than or equal to 10)

case 2 x - 10 < 0, so x < 10
x - 16 = -6
x = 4(accepted) since x has to be less than 10

Is this approach correct? ### That approach works, but it

That approach works, but it seems more time-consuming than the approach taught in https://www.gmatprepnow.com/module/gmat-algebra-and-equation-solving/vid...

### For this kind of question, I

For this kind of question, I assume that the only way to find which statement would be to solve both equations as is shown in the video? Isn't that time consuming? I'm just worried about having to do those calculations if it takes tons of time. ### I don't believe there's a

I don't believe there's a faster solution than the one shown on the video.

This question is, indeed, time-consuming, but I think it can be done in under 2 minutes. Keep practicing!

Cheers,
Brent

### Question: can't you just

Question: can't you just notice that the solution 4 is common to both statement 1 and 2, and know that it is the answer? do you need to re-plug all the solutions back into the equation every time? ### Just because 4 is a POSSIBLE

Just because 4 is a POSSIBLE solution to an equation doesn't mean it IS a solution. You must verify it.
That said, I do admit that extraneous roots are not very common on the GMAT, but it's always safe to test the solutions.

### I have come across the

I have come across the following statements on DS questions:

1. |a| = a
2. |a| = -a

What does it tell us? ### 1. |a| = a

1. |a| = a
Since |a| always greater than or equal to 0, this statement tells us that a ≥ 0

2. |a| = -a
Since |a| always greater than or equal to 0, this statement tells us that -a ≥ 0
Now take -a ≥ 0 and multiply both sides by -1 to get a ≤ 0

When we COMBINE the two inequalities, we get: a ≤ 0 ≤ a, which means a MUST equal 0

Cheers,
Brent

### https://gmatclub.com/forum/if

https://gmatclub.com/forum/if-denotes-the-greatest-integer-less-than-or-equal-to-z-and-162078.html

I don't understand this question. I tried to make sense of it by reading it as whatever lzl = to that number is less than or equal to z so if lzl = -1 then I know that -1 must be less than or equal to -1. I can't comprehend the rest. Same with this question:

https://gmatclub.com/forum/if-denotes-the-least-integer-greater-than-or-equal-to-x-132223.html ### Here are my solutions to your

Here are my solutions to your two questions.

Cheers,
Brent

### Hi Brent,

Hi Brent,

I thought for the purpose of the test, we don't consider root(n) to be both a positive and a negative number. For e.g. sqrt 169 can only be 13 or can it also be -13?

I am using this in the question here https://gmatclub.com/forum/what-is-the-value-of-x-100526.html

If i consider both the roots, the answer is C. If I don't, it should be B. Am i making another one of my careless mistakes? :)

Thank you! Be careful. The equation k² = 169 is not the same as the equation k = √169

The main idea here is that the square root NOTATION (√) directs us to single out the POSITIVE square root of a value.
So, for example, √4 = 2 (not -2)

Conversely, the equation k² = 169 does not contain any square root NOTATION.
As such, there are two possible solutions: k = 13 and k = -13

In general, we can say that, if k² = q (where q > 0), then EITHER k = √q OR k = -√q

We can also say that, if k = √n (where n > 0), then there is only ONE possible value of k

Does that help?

Cheers,
Brent

### Yes, definitely. Thank you!

Yes, definitely. Thank you!

### Hi Brent,

Hi Brent,
Could you please solve the following question? I couldn't understand OG solution.

If q, s and t are all different numbers, is q<s<t?
(1) t-q = |t-s| + |s-q|
(2) t>q

OG 2019 - DS07441 ### Hi Brent!

Hi Brent!

After we solve for both positive & negative values in Statement 2, |x^2 - 16| = x - 4, and identify that x = 4 or x = -3 in positive value scenario and x = 4 or x = -5 in negative value scenario; are we able to just skip over the step where we plug in each x value (-3, -5, and 4) into the original Statement 2 equation and conclude that x = 4 since x = 4 shows up both times as a possible solution in both positive and negative scenarios for Statement 2?

Just trying to identify any ways to save extra time on these types of questions on exam day, if possible.

*By positive & negative scenarios for Statement 2 |x^2 - 16| = x - 4, I'm referring to:
Positive: x^2 - 16 = x - 4
Negative: x^2 - 16 = -x + 4 ### Great question, Nicole!

Great question, Nicole!
If the same solution appears in both the positive and negative scenarios, we we still need to test the solutions.

Consider this equation: |x² - 16| = x + 4
In this case, the positive scenario yields the solutions x = -4 and x = 5
And the negative scenario yields the solutions x = -4 and x = 3

However, for this equation, all three solutions (x = -4, x = 5 and x = 3) all satisfy the equation.
So, we must test each possible solution.

Cheers,
Brent

### Hi Brent,

Hi Brent,

In statement 2, after having 2 equation
x^2 - x - 12 = 0
x^2 + x - 20 = 0
Could I use the elimination method in this case ?

Then I gonna have the following result:
2x^2 - 32 = 0
x^2 = 16
x = 4.
Then I don't have to check the extraneous root :D ### That's a good idea.

That's a good idea. Unfortunately, it will yield intended results.

Take, for example, the following equation: |x²| = x
This equation has two possible solutions: x = 1 and x = 0

Now let's see what happens if we try to apply your elimination method.

If we take |x²| = x, the two resulting equations are:
x² = x
x² = -x

From here, if we subtract the bottom equation from the top equation, we get: 0 = 2x, which means the solution is x = 0
Notice that the elimination method ELIMINATED one of the answer choices (x = 1)

In your solution, something very different happens.
In your solution, you got: x² = 16
From there, you solved the equation to get x = 4.
However, the equation x² = 16 actually has TWO solutions: x = 4 and x = -4
While x = 4 IS a solution to the given equation, x = -4 is NOT a solution.

So, in my first example, the elimination method ELIMINATED a possible solutions.
In the second example, the elimination method CREATED a false solution.

For these reasons, you should avoid using that strategy.