How do you solve #-x^2-9x-16<=-2#?

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Answer 1 · 2017-09-02T12:44:47

#x >=-2 or x >=-7#

#-x^2 - 9x - 16 <= - 2#

According to the law of inequality, when you multiply through with minus #(-)# sign, the inequality sign changes..

#-x^2 - 9x - 16 <= - 2#

Multiply through by #(-)#

#-(-x^2 - 9x - 16) <= -(- 2)#

Note #- xx - = +#

#x^2 + 9x + 16 >= 2#, Note that the inequality sign changes..

Collect like terms

#x^2 + 9x + 16 - 2 >= 0#

#x^2 + 9x + 14>= 0#

**Solving the Quadratic Equation, we have #7 and 2# as the factors..

Where #-> 7 xx 2 = 14, 7x + 2x = 9x#

Hence we have..

#x^2 + 9x + 14>= 0 rArr x^2 +2x +7x + 14 >=0#

#x^2 +2x +7x + 14 >=0#

#(x^2 +2x) (+7x + 14) >=0#

#x(x +2) +7 (x + 2) >=0#

#(x +2) (x + 7) >=0#

#x +2 >=0 or x + 7 >=0#

#x >=-2 or x >=-7#