S = { x: x^2 + 2x - 8 < 0}

x^2 + 2x - 8 < 0

Let us factor:

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

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

==> x < -4 and x > 2 (impossible)

OR:

(x+4) > 0 and (x-2)< 0

==> x > -4 and x < 2

==> -4< x < 2

==> x belongs to (-4, 2)

==> S = { x : x belongs to (-4, 2)}

To determine the elements of the set, we'll sove the inequality.

For this purpose, will solve first the equation:

x^2 + 2x - 8= 0

We'll apply the quadratic formula:

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

x1 = (-2+sqrt36)/2

x1 = (-2+6)/2

x1 = 2

x2 = (-2-6)/2

x2 = -4

Now, we could write the inequality as:

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

For a product of 2 factors to be negative, one factor must be positive and the other must be negative.

We'll consider 2 cases:

**Case 1:**

(x-2)<0

x<2

(x+4)>0

x>-4

The solution of this case includes all numbers between -4 and 2, but not the numbers themselves.

x belongs to the interval (-4,2)

**Case 2:**

(x-2)>0

x>2

(x+4)<0

x<-4

There is not solution for this case.

Therefore, the required set is:

**S = {x/-4 < x < 2}**

x^2+2x-8 <0. To find the set S to which x belongs.x.

Solution:

If x1 and x2 are the roots of x^2+x-8, then x^2+2x-8 < o for x belonginging to the interval {x1, x2} as x^2+2x-8 = (x-x1)(x-x2), the factors has opposite sign .

So we factorise x^2+2x-8 = 0

(x^2+2x+1) -9 = 0

(x+1)^2 -3^2 = 0.

(x+1+3)(x+1-3) = 0

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

x+4 = 0 or x-2 = 0

x1 = -4 and x2 = 2

So for x belonging to (-4 , 2) , expression x^2+2x-8 <0 .