How do you solve #-x^2-4x>3#?

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How do you solve #-x^2-4x>3#?

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Answer 1 · 2015-09-27T17:15:53

#-3 < x <-1#

Explanation:

#-x^2-4x>3=># multiply by -1, reverse direction of inequality:
#x^2+4x<-3=># solving by completing the square:
#x^2+4x+4<4-3#
#(x+2)^2<1#
#-1 < (x+2)<1#
#-3 < x <-1#
In interval form:
#(-3, -1)#

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How do you solve #-x^2-4x>3#?

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