If |(x 3)^2 + 2|

Question

If  |(x – 3)^2 + 2| < |x – 7|  , which of the following expresses the allowable range for x?

(A) 1 < x < 4

(B) 1 < x < 7

(C) – 1 < x < 4 and 7 < x

(D) x < – 1 and 4 < x < 7

(E) – 7 < x < 4 and 7 < x

Absolute value inequalities are a rare and tricky topic on the GMAT. For a detailed discussion of this question type, as well as the OE for this particular question, see:
Absolute Value Inequalities

Mike

amulya619 · Accepted Answer

Hi,

I always go for plugging in for these kind of questions. Its fast and quite reliable.

Lets look at one option at at time:-

A. 1<x<4: Lets take X= 2 and 3, the equation is satisfied. Hold on to it.

B. 1<x<7: We have already tested the equation for x=2 and 3. Now, we can check it with X=5, which gives the result as false. So, discard this option.

C. -1<x<4 and x<7: In the above statement, we have already tested for X=5 or we can say that for the second part of this statement x<7. So, this statement is false. However, just to be sure, putting x=0 will give false. So, discard.

D. X<-1 and 4<x<7: As done earlier, the statement has proven to be false for X=5. So, this statement can also be discarded.

E. -7<x<4 and x<7: Already tested for X=0, 2 and 5. So discard.

Thus the correct answer is A.

broall · Answer

Note that (x-3)^2+2 \geq 2 >0 \quad\forall x \implies |(x – 3)^2 + 2| = (x – 3)^2 + 2 If x \geq 7 \implies (x – 3)^2 + 2 < x – 7 \implies x^2 -6x +9+20 So in this case, there is no x satisfied the equation. If x <7 \implies (x – 3)^2 + 2 < 7-x \implies x^2 -6x +9+2<7-x \implies x^2 -5x +4 <0 \implies (x-1)(x-4)<0 \implies 1

nkin · Answer

I looked at the options and substituted 6 and saw anything greater than 6 was also not going to work and was able to eliminate B,C,D,E in one shot.

Kem12 · Answer

Yes the answer may seem correct but I don't understand how (x−1)(x−4)<0⟹1<x<4⟹(x−1)(x−4)<0⟹1<x<4 
Shouldn't it be x< 1 and x<4 and the correct answer just x<4 It looks like a subset.

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gmatzpractice · Answer

Based on Answer Choices, try x=5 
Inequality not true 
Eliminate B, C, D, E 
Answer A

Nistha86 · Answer

Hi Experts can you please show the result not by plugging in values but solving the inequality?

GMATBUDDING · Answer

@bunuel@karishma

Do we have any other ways apart from plugin of values ?

HoudaSR · Answer

The quickest method to solve these type of questions considering that you are given ranges in the answer choices is  to plugging values and proceed by elimination.

But if you want to solve algebraically,  the algebraic method I used is as following:

|(x – 3)^2 + 2| < |x – 7| is equivalent to the below (remember that |x| = sqrt (x)^2) :

sqrt ((x-3)^2 +2)^2 < sqrt (x-7)^2
Raise to the power 2 to get rid of the square root: (x-3)^2 +2)^2 < (x-7)^2
Move everything to the LHS: (x-3)^2 +2)^2 - (x-7)^2 < 0 (this is a quadratic in the form a^2-b^2= (a-b) (a+b) )
Simplifying further, we get: ((x-3)^2 +2 -x+7) ((x-3)^2 +2+x-7) <0
Which finally simplifies to: (x^2-5x+4) (x^2-7x+18) < 0

The first quadratic x^2-5x+4 can be written as (x-4) (x-1)

The second quadratic x^2-7x+18 has no roots because the desciminant b^2-4ac<0 
When a quadratic has no roots that means that the prabola of it on the coordinate plane is either above the x axis or below the x axis, but it doesn't cut the x axis at all.
And by setting x=0 on x^2-7x+18 we get y=18 and that is positive; therefore x^2-7x+18 will always be positive (above the y axis).

Now going back to our inequality: (x^2-5x+4) (x^2-7x+18) < 0
We can now rewrite it as: (x-4) (x-1) (x^2-7x+18) <0
And since x^2-7x+18 is always positive, we need to determine the range of x for when (x-4) (x-1) is negative. 
Drawing that on the number line we get   ........+........1............-.............4...........+........

Therefore the allowable range for x is 1<x<4

And the correct answer is A.

bumpbot · Answer

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If |(x 3)^2 + 2| < |x 7| , which of the following expresses the al

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