Science

Math

Humanities

... and beyond

Socratic Meta
Featured Answers

Topics

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

1 Answer

Mar 2, 2017

#x < -8 or x > -4#

Explanation:

#-x^2 - 12 x < 32#
minus with 32,

#-x^2 - 12 x -32 < 0# or #x^2+12 x +32 > 0#

Factorized,

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

                         -oo,              -8,           -4,              +oo

 (-x-8)                               +              -                 -

(x + 4)                               -              -                 +

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

Therefore,
#x < -8 or x > -4#

Related questions

See all questions in Polynomial Inequalities

Impact of this question

996 views around the world

You can reuse this answer
Creative Commons License