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

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hala718 | High School Teacher | (Level 1) Educator Emeritus

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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).

 

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utkr940 | Student, Grade 11 | (Level 1) Honors

Posted on

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

 

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

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.)

thewriter's profile pic

thewriter | College Teacher | (Level 1) Valedictorian

Posted on

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.

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neela | High School Teacher | (Level 3) Valedictorian

Posted on

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).

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