fortsill wrote:
Fig wrote:
C. x + 2 < |x2 - 4|
|x2 - 4| = |(x+2)(x-2)|
Since the sign of a*x^2+b*x+c is that of a for any values of x not between the 2 roots, solutions of the equation a*x^2+b*x+c = 0 if they exist.
Thus,
o Sign((x+2)(x-2)) > 0 when x > 2 or x < -2
o Sign((x+2)(x-2)) < 0 when -2 < x < 2
Trying hard to grasp this concept, but don't think I'm understanding what you're trying to convey here.
Ideally, I would try to solve this as such:
Given that |x^2 - 4| is same as |(x+2)(x-2)|
Solving for x in
|x^2 - 4| > x + 2 in the following two scenarios:
1. When (x+2)(x-2) >= 0
==> x^2 - 4 > x + 2
==> x>-1 and x>2
2. When (x+2)(x-2) < 0
==> -(x^2 - 4) > x + 2
==> 4 - x^2 > x + 2
==> x<1 and x<-2
Does the combination of these solutions for x represent the final answer? Am I missing a step?
First of all you won't see such equation on the GMAT, but anyway:
\(|x^2-4|>x+2\).
We should consider the cases when \(x^2-4<0\) and \(x^2-4\geq{0}\):
\(x^2-4<0\) for \(-2<x<2\) --> \(-(x^2-4)>x+2\) --> \(x^2+x-2<0\) --> \((x+2)(x-1)<0\)--> \(-2<x<1\). Since we are considering the range \(-2<x<2\) then the solution for this case will be intersections of these two (common part): \(-2<x<1\);
\(x^2-4\geq{0}\) for \(x\leq{-2}\) or \(x\geq{2}\) --> \((x^2-4)>x+2\) --> \(x^2-x-6>0\) --> \((x+2)(x-3)>0\)--> \(x<-2\) or \(x>3\). Since we are considering the range \(x\leq{-2}\) or \(x\geq{2}\) then the solution for this case will be intersections of these four (common parts): \(x<-2\) or \(x>3\);
Finally the ranges of \(x\) for which \(|x^2-4|>x+2\) true are: \(x<-2\), \(-2<x<1\) and \(x>3\).
Solving inequalities:
x2-4x-94661.html#p731476inequalities-trick-91482.htmldata-suff-inequalities-109078.htmlrange-for-variable-x-in-a-given-inequality-109468.html?hilit=extreme#p873535everything-is-less-than-zero-108884.html?hilit=extreme#p868863Hope it helps.