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

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.

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

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

@bunuel@karishma

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

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.