Question 502207
{{{abs(x^2+6x) = 3x+18}}} Remove the absolute-value bars from the left side and you get two equations thus:
1) {{{x^2+6x = 3x+18}}} or 2) {{{x^2+6x = -(3x+18)}}} Now solve each of these equations for x.
1) {{{x^2+6x = 3x+18}}} Rewrite as a quadratic equation in standard form:
{{{x^2+3x-18 = 0}}} Factor this quadratic equation.
{{{(x-3)(x+6) = 6}}} Apply the "zero product" rule:
{{{x-3 = 0}}} or {{{x+6 = 0}}} so that...
{{{highlight(x = 3)}}} or {{{highlight(x = -6)}}}
2) {{{x^2+6x = -(3x+18)}}} Simplify the right side.
{{{x^2+6x = -3x-18}}} Rewrite as a quadratic equation in standard form:
{{{x^2+9x+18 = 0}}} Factor this quadratic equation.
{{{(x+3)(x+6) = 0}}} Apply the "zero product" rule:
{{{x+3 = 0}}} or {{{x+6 = 0}}} so that...
{{{highlight(x = -3)}}} or {{{highlight(x = -6)}}}