# What is x if E(x)=(x^2-4)/(x^2+2x-35) is positive?

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To determine x if E(x) = (x^2-4)/(x^2+2x-35).

Solution:

Given E(x) = (x+2)(x-1)/{(x+7)(x-5)} > 0.

We know that -7< -2< 1< 5.

So E(x) = (x+2)(x-1)/(x+7)(x-5) > 0 when **x>5** as all factors are are positive.

When** -2< x< 1**, E(x) is positive as both numerator and denominators are negative. So E(x) is positive.

When** x< -7**, all factors are negative. So E(x) > 0.

**Therefore when E(x) > 0 for x< -7 or -2< x < -1 or x> 5 .**

For the given expression to be positive, the numerator and denominator have to be both positive or both negative:

We'll verify where the numerator is positive:

x^2 - 4 = 0

(x-2)(x+2) = 0

x1 = 2 and x2 = -2

The numerator is negative over the range (-2,2) and it is positive over the reunion of ranges (-infinite,-2)U(2;+infinite).

We'll discuss the sign of denominator.

x^2+2x-35 = 0

We'll apply the quadratic formula:

x1 = [-2+sqrt(4 + 140)]/2

x1 = 5

x2 = -7

The denominator is positive over the reunion of ranges (-inf.,-7)U(5,+inf.) and it is negative over the range (-7;5).

The intersection between intervals where the numerator and denominator are both positive or both negative is:

**E(x)>0 <=> x is in the ranges (-inf.;-7)U(-2;5)U(2;+inf.).**