Question 224569
Given:{{{sqrt(1-3x) = x+1}}} solve for x:
{{{sqrt(1-3x) = x+1}}} Square both sides of the equation.
{{{(sqrt(1-3x))^2 = (x+1)^2}}} 
{{{1-3x = x^2+2x+1}}} Add 3x to both sides.
{{{1-3x+3x = x^2+2x+3x+1}}} Simplify.
{{{1 = x^2+5x+1}}} Subtract 1 from both sides.
{{{0 = x^2+5x}}} Factor an x from the right side.
{{{0 = x(x+5)}}} Apply the "zero product" rule.
{{{x = 0}}} or {{{x+5 = 0}}}
If {{{x+5 = 0}}} then {{{x = -5}}}
So the solutions are:
{{{x = 0}}}
{{{x = -5}}}