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 < x

Note that \((x-3)^2+2 \geq 2 >0 \quad\forall x \implies |(x – 3)^2 + 2| = (x – 3)^2 + 2\)

If \(x \geq 7 \implies (x – 3)^2 + 2 < x – 7 \implies x^2 -6x +9+2<x-7 \implies x^2 -7x +18 <0\)

We have \(x^2 -7x+18 = x^2 -2 \times x \times \frac{7}{2} +(\frac{7}{2})^2 +\frac{23}{4} = (x-\frac{7}{2})^2 +\frac{23}{4} >0\)

So in this case, there is no \(x\) satisfied the equation.

If \(x <7 \implies (x – 3)^2 + 2 < 7-x \implies x^2 -6x +9+2<7-x \implies x^2 -5x +4 <0\)

\(\implies (x-1)(x-4)<0 \implies 1<x<4\)

The answer is A

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.