# For what values of x is the following inequation valid :  2x^2+ 14x + 24 > 0

william1941 | Student

We are given the following inequation to solve 2x^2+ 14x + 24 > 0

Now, 2x^2+ 14x + 24 > 0

=> 2x^2 + 6x + 8x +24 >0

=> 2x ( x +3) + 8 (x+3) >0

=> (x+4)(x+3)>0

Now as (x+4)(x+3) is greater than 0. Either (x+4)and (x+3)can be greater than 0 or they can both be less than 0.

If they are greater than 0,

=> (x+4) > 0 and (x+3) >0

=> x > -4 and x >-3

=> x has to be greater than -3

If both (x+4)and (x+3) are less than 0

=> (x+4) <0 and (x+3) <0

=> x < -4 and x <-3

=> x has to be less than -4

Therefore x can be either greater than -3 or less than -4

neela | Student

2x^2+14x+24 > 0

To find the values of x  the inequality is invalid.

We first find the roots of the equation 2x^2+14x+24 =0.

We divide the equation by 2:

x^2+7x+12 = 0

x^2+4x+3x+12 = 0

x(x+4) + 3(x+4) = 0

(x+4)(x+3) = 0

The roots are -4 and -3.

-4 < -3.

So the expression (x+4)(x+3) < 0 for the given inequality to be invalid.

So  (+4)(x+3) < 0 when   -4  < x < -3. Or x lies between the roots of the expression.

So  2x^2+14x +24 > 0 is invalid when (x+4)(x+3) < 0 ) or when -4 < x <-3.