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 $x_4=3$

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