How do you solve #x^2+16x+24>6x#?

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Answer 1 · 2018-05-25T08:10:13

#x<-6#or #x>-4#

simplifying to #x^2+10x+24>0#

solving #x^2+10x+24=0# we get
#x_1=-4#
#x_2=-6#
so our inequality is equivalent to
#(x+4)(x+6)>0#
this gives us
#x>-4# or #x<-6#

Answer 2 · 2018-08-09T22:50:42

#x in (-oo, -6) uu (-4, oo)#

Given:

#x^2+16x+24 > 6x#

Subtract #6x# from both sides to get:

#x^2+10x+24 > 0#

We can make the left hand side into a perfect square trinomial by adding #1#, so let us add #1# to both sides to get:

#(x+5)^2 = x^2+10x+25 > 1#

Note that this would give equality when #(x+5)^2 = 1#, i.e. when #x+5 = +-1#, i.e. when #x=-6# or #x=-4#.

Hence the inequality is achieved when:

#x < -6" "# or #" "x > -4#

In interval notation, when:

#x in (-oo, -6) uu (-4, oo)#