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Value of x in a quadratic inequality

Expert replies
by Brent@GMATPrepNow » Sat Feb 11, 2017 6:41 am
Here's a question I just made up.
Difficulty level: 650 - 700
If x is an integer, what is the value of x?

(1) x² + 8x + 16 > 1
(2) x² - 8x + 9 < -6
Answer: B

Cheers,
Brent
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Source: — Data Sufficiency |

by DavidG@VeritasPrep » Sat Feb 11, 2017 8:35 am
Brent@GMATPrepNow wrote:Here's a question I just made up.
Difficulty level: 650 - 700
If x is an integer, what is the value of x?

(1) x² + 8x + 16 > 1
(2) x² - 8x + 9 < -6
Answer: B

Cheers,
Brent
Brent original!

The first statement, you can pretty much just eyeball and see it's not sufficient alone. x= 0 will work. So will x = 1. Clearly not sufficient

Statement 2 is more interesting. (Advice: anytime one statement can be dismissed easily and the other is significantly more difficult to evaluate, slow down. We can simplify to x^2 - 8x + 15 < 0, which becomes (x -5)(x - 3) < 0. If we're multiplying two elements together to get a negative product, we know that one element is positive and one is negative. Because x - 3 is larger than x - 5, x - 3 must be the positive value and x - 5 should be the negative value. You could pick numbers and show that only x = 4 will work.

Algebraically, if x - 3> 0, then x > 3. If x - 5 < 0, then x < 5. If 3 < x < 5, x must be 4. Sufficient. The answer is B

(And notice how important it is that we're dealing with an integer value!)

Nice question.
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by Brent@GMATPrepNow » Sun Feb 12, 2017 9:37 am
Brent@GMATPrepNow wrote:If x is an integer, what is the value of x?

(1) x² + 8x + 16 > 1
(2) x² - 8x + 9 < -6
Target question: What is the value of x?

Statement 1: x² + 8x + 16 > 1
Subtract 1 from both sides to get: x² + 8x + 15 > 0
As David pointed out, we might see right away that MANY different values of x will satisfy this inequality.
To begin, any POSITIVE value of x will satisfy the inequality x² + 8x + 15 > 0
Since we cannot answer the target question with certainty, statement 1 is NOT SUFFICIENT

Statement 2: x² - 8x + 9 < -6
Add 6 to both sides to get: x² - 8x + 15 < 0
Factor to get: (x - 3)(x - 5) < 0

With quadratic inequalities like this, it's useful to first examine and solve the corresponding EQUATION
So, solve: (x - 3)(x - 5) = 0
We get: x = 3 or x = 5
These are our CRITICAL VALUES of x (i.e., x-values that satisfy the corresponding EQUATION)

Now let's examine what happens within the x-values LESS THAN and GREATER THAN these critical values.
There are 3 ranges to consider:
i) x is less than 3
ii) x is between 3 and 5
iii) x is greater than 5

i) x is less than 3
If x < 3, then (x - 3) is NEGATIVE and (x - 5) is NEGATIVE
So, (x - 3)(x - 5) = (NEGATIVE)(NEGATIVE) = POSITIVE
So, when x is less than 3, (x - 3)(x - 5) > 0
We're looking for x-values that satisfy the inequality (x - 3)(x - 5) < 0, so we can conclude that x is NOT less than 3

ii) x is between 3 and 5
If 3 < x < 5, then (x - 3) is POSTIVE and (x - 5) is NEGATIVE
So, (x + 3)(x + 5) = (POSITIVE)(NEGATIVE) = NEGATIVE
So, when x is between 3 and 5, (x - 3)(x - 5) < 0
Perfect, we're looking for x-values that satisfy the inequality (x - 3)(x - 5) < 0!
So we can conclude that x IS between 3 and 5

iii) x is greater than 5
If x > 5, then (x - 3) is POSITIVE and (x - 5) is POSITIVE
So, (x - 3)(x - 5) = (POSITIVE)(POSITIVE) = POSITIVE
So, x is greater than 5, (x - 3)(x - 5) > 0
We're looking for x-values that satisfy the inequality (x - 3)(x - 5) < 0, so we can conclude that x is NOT greater than 5

We have seen that x IS between 3 and 5
Since 4 is the ONLY integer between 3 and 5, we can be certain that x = 4
Since we can answer the target question with certainty, statement 2 is SUFFICIENT


Answer: B

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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