# Question: Quadratic and Absolute Value

## Comment on Quadratic and Absolute Value

### Hi Brent, first of all thank

Hi Brent, first of all thank you for all the lessons, I am having so much fun preparing for GMAT now.

In this question, in the last step of the second statement you got:

3 <2x <5

After this step you divide all the sides by 2. From the earlier lessons, I have gathered that when we divide inequality the signs change in the reverse direction.

So shouldn't it be 1.5 > x > 2.5 as the final answer. The statement still will be sufficient. But just wanted a clarity on the rules.

Thank you Brent. ### Be careful. There are two

Be careful. There are two rules pertaining to this.

If we divide or multiply both/all sides of an inequality by a POSITIVE value, then the direction of the inequality signs stay the SAME.

If we divide or multiply both/all sides of an inequality by a NEGATIVE value, then the direction of the inequality signs are REVERSED.

### Are there any other methods

Are there any other methods solve the first statement other than checking on the number line? ### The number line isn't 100%

The number line isn't 100% necessary for solving these questions. Once you've factored the quadratic and determined the critical points (the values of x such that the quadratic evaluates to be 0), you can just test various values, without plotting the outputs on the number line.

### I do not understand statement

I do not understand statement 2. |2x-4|<1

So shouldnt that be

1) 2x - 4 < 1 -------- x < 2.5
2) 2x - 4 < -1 -------- 2x < 3 ------x < 1.5

Why do you have x > 1.5? Because the negative sign should only go to the right hand side when you open the | | of the left hand side. Why are you multiplying the left hand side by negative also? thanks ### YOUR QUESTION: Why are you

YOUR QUESTION: Why are you multiplying the left hand side by negative also?
I'm not multiplying anything. I'm applying a rule regarding absolute value inequalities, which goes like this:

When solving inequalities involving ABSOLUTE VALUE, there are 2 things you need to know:

- Rule #1: If |something| < k, then –k < something < k
- Rule #2: If |something| > k, then EITHER something > k OR something < -k
(Note: these rules assume that k is positive)

This is covered in the following video: https://www.gmatprepnow.com/module/gmat-algebra-and-equation-solving/vid...

GIVEN: |2x - 4| < 1
From Rule #1, we get: -1 < 2x - 4 < 1
----------------------------

Please note that we can test your theory by plugging in numbers.

For example, if |2x - 4| < 1, then one possible solution is x = 2
When we plug in x = 2, we get: |2(2) - 4| < 1
This simplifies to be |0| < 1, which is true.

You wrote: 2) 2x-4 < -1 ----- 2x < 3 ---- x < 1.5
This tells us that x must be less than 1.5, however we just showed that x = 2 is a possible solution.

Does that help?

Cheers,
Brent

### Thanks. Is there a similar

Thanks. Is there a similar rule for squares in inequalities? I mean is that the same rule if it was |2x-4|^2? Or is it different when we deal with squares? thanks ### There's no general rule; you

There's no general rule; you'll have to approach that format on a case-by-case basis.
That said, it would be VERY RARE to encounter a GMAT question involving the square of an absolute value.

Cheers,
Brent

### Actually my question is

Actually my question is different. I will rephrase. If there we are dealing with squares with inequalities, is there a rule that we can use just like you shared the rule for absolute values? thanks ### The rules/strategies won't be

The rules/strategies won't be the same.

Consider these examples:
If x² < 9, then we can write: -3 < x < 3
If x² < 49, then we can write: -7 < x < 7
If x² < 100, then we can write: -10 < x < 10

IN GENERAL: If x² < k (where k ≥ 0), then we can write: -√k < x < √k
-------------------------------------

What about when the square is greater than some value?
Consider these examples:
If x² > 9, then we can write: x > 3 or x < -3
If x² > 49, then we can write: x > 7 or x < -7
If x² > 121, then we can write: x > 11 or x < -11

IN GENERAL: If x² > k (where k ≥ 0), then we can write: x > √k or x < -√k

Does that help?

Cheers,
Brent

Thanks so much!

### Hi Brent, what if there were

Hi Brent, what if there were more than one value in the middle region? Would the 1st statement be sufficient in that case ? ### If there were MORE THAN ONE

If there were MORE THAN ONE integer value within the given range, then statement 1 would be insufficient, since there would be more than one possible value for x.

### both statements are identical

both statements are identical in their solution, we can solve both equations and get only one integer satisfied which is 2 That's correct.

### Hi Brent, for St 1, since I

Hi Brent, for St 1, since I just regard it as normal Quadratic equation and arrive at x= 1 or 3, therefore insufficient. How will I know when to resolve it as quadrtic as it is or as number line way? Thanks Brent Be careful; the quadratic EQUATION x² - 4x + 3 = 0 is different from the quadratic INEQUALITY x² - 4x + 3 < 0.

In the EQUATION, we are saying that x² - 4x + 3 EQUALS 0.
In the INEQUALITY, we are saying that x² - 4x + 3 does NOT equal 0. Instead, we are saying that x² - 4x + 3 < 0.

### Thanks Brent for the

Thanks Brent for the clarification. So number line for inequality quadratic equation. Noted thanks!! That's correct.