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Test x=0 in |(x–3)² + 2|<|x–7|:Mo2men 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
Can you elaborate more on first case? how do you find no solution?flprg wrote: Then, you split your equation in 2 parts : when x>7 and x<7
1st case : x^2-6x+11<x-7 which gives x^2-7x+18<0 ==> No solution since x^2-7x+18 is always positive.
Dear Mitch,GMATGuruNY wrote:Test x=0 in |(x–3)² + 2|<|x–7|:Mo2men 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
|(0–3)² + 2|<|0–7|
11 < 7.
Does not work.
Eliminate any answer choice with a range that includes x=0.
Eliminate C and E.
Test x=5 in |(x–3)² + 2|<|x–7|:
|(5–3)² + 2|<|5–7|
6 < 2.
Does not work.
Eliminate any remaining answer choice with a range that includes x=5.
Eliminate B and D.
The correct answer is A.
That's from your high school maths classes. You calculate the discriminant D = b^2-4ac (here it is (-7)^2-4*1*18). When D is negative, your quadratic expression has the sign of a (the coefficient of x^2). Here a = 1 so x^2-6x+11 is always positive, thus no solution to the equation above. Yes, I admit it is more brutal if your not familiar with these concepts.Mo2men wrote:Can you elaborate more on first case? how do you find no solution?flprg wrote: Then, you split your equation in 2 parts : when x>7 and x<7
1st case : x^2-6x+11<x-7 which gives x^2-7x+18<0 ==> No solution since x^2-7x+18 is always positive.
I tried it by plugging the answer choicesMo2men 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
Source: Magoosh
The CRITICAL POINTS are the values of x such that the two sides of the inequality are EQUAL.Mo2men wrote:
(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
Dear Mitch,
Does the critical point method work here> if yes, what are those points??
