How do you solve #abs(x^2+6x)= 3x+18#?

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Answer 1 · 2018-06-08T08:31:38

#x_1=x_2=-6#, #x_3=-3# and #x_4=3#

#|x^2+6x|=3x+18#

#(x^2+6x)^2=(3x+18)^2#

#(x*(x+6))^2=(3*(x+6))^2#

#x^2*(x+6)^2=9*(x+6)^2#

#(x^2-9)*(x+6)^2=0#

#(x+6)^2(x+3)(x-3)=0#

So #x_1=x_2=-6#, #x_3=-3# and #x_4=3#