Question

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

Accepted Answer

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).

Answer

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

Answer

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.)

Answer

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.

Answer

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).

Math

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

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