# x^2 + 2x -8 > 0 find x values.x^2 + 2x -8 > 0 find x values.

### 5 Answers | Add Yours

x^2 + 2x - 8 > 0

First let us facotor:

x^2 + 2x -8 = (x+4)(x-2) > 0

Then we have two cases:

In order for the function to be positive, then :

(x+4) > 0 AND (x-2) > 0

==> x > -4 and x > 2

==>** x belongs to (2,inf)...........(1)**

The second case is:

(x+4)< 0 AND (x-2) < 0

==> x < -4 and x < 2

==> **x belongs to (-inf, -4)...........(2)**

**==> from (1) and (2) we conclude that:**

**x belongs to (-inf, -4) U (2,inf).**

x^2+2x-8>0

therefore, x^2+4x-2x-8>0

=x(x+4)-2(x+4)>0

=(x+4)(x-2)>0

therefore, the values of x can be as follows:-

case 1. if both (x+4) and (x-2)>0.

x+4>0

therefore, x>-4

and x-2>0

therefore,x>2

case 2. if both (x+4) and (x-2)<0

x+4<0

therefore, x<-4

x-2<0

therefore, x<2

To find where the expression is positive, we'll calculate first the roots of the equation:

x^2 + 2x -8 = 0

We'll apply the quadratic formula:

x1 = [-2+sqrt(4+32)]/2

x1 = (-2+6)/2

**x1 = 2**

x2 = (-6-2)/2

**x2 = -4**

The expression is negative between the roots and positive outside the roots.

**So, the intervals where the expression is positive are:**

**(-inf., -4) U (2 , +inf.)**

To solve the given inequality we can use the following steps.

x^2+4x-2x-8>0

=>x(x+4)-2(x+4)>0

=>(x-2)(x+4)>0

Now (x-2)(x+4)>0 if either both (x-2) and (x+4) are less than 0 or both (x-2) and (x+4) are greater than 0.

This means either x-2>0 or x>2 and x+4>0 or x>-4 therefore x>2.

If both (x-2) and (x+4) are less than 0. It means x<2 and x<-4. Therefore x<-4.

From the two results we get that all values of x such that x>2 and x<-4 satisfy the given relation.

To solve x^2+2x-8 > 0.

Solution:

This is an inequality . And a quadratic inequality. We solve it similar to the equality for its roots x1 and x2. Then,

x^2+2x-8 = (x-x1)(x-x2).

Now the variable x should lie between the 2 roots x1 and x2 so that the one of the factor in the product (x-x1)((x2-x) is negative and the product value becomes negative or less than zero.

So we factorise x^2+2x-8 :

x^2+2x -8 = x^2+ 4x-2x-8

= x(x+4) -2(x+4)

So x^2+2x-8 = (x+4)(x-2).

The roots (or zeros of x^2+2x-8) are x1=-4 and x2 =2.

So

Therefore x^2+2x-8 = (x+4)(x-2) becomes lessthan zero or negative iff x has value in betwee -4 and 2.

So x^2+2x-8 = (x+4)(x-2) < 0 for x in the interval (-4 , 2).