How do you solve -x^2-12x<32−x2−12x<32 by algebraically?

1 Answer
salamat
Mar 2, 2017

x < -8 or x > -4x<−8orx>−4

Explanation:

-x^2 - 12 x < 32−x2−12x<32
minus with 32,

-x^2 - 12 x -32 < 0−x2−12x−32<0 or x^2+12 x +32 > 0x2+12x+32>0

Factorized,

(-x - 8 )(x + 4 ) < 0(−x−8)(x+4)<0 or (x+8)(x+4) > 0(x+8)(x+4)>0

                         -oo,              -8,           -4,              +oo

 (-x-8)                               +              -                 -

(x + 4)                               -              -                 +

(-x-8)(x+4)                        -              +                 -

Therefore,
x < -8 or x > -4x<−8orx>−4

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