mikemcgarry wrote:
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 < xAbsolute 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 InequalitiesMike
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.