# How do you solve  3x^2 = -9x?

Apr 7, 2018

The two solutions are $x = 0$ and $x = - 3$.

#### Explanation:

First, bring everything to one side of the equation:

$3 {x}^{2} = - 9 x$

$3 {x}^{2} \textcolor{b l u e}{+} \textcolor{b l u e}{9 x} = - 9 x \textcolor{b l u e}{+} \textcolor{b l u e}{9 x}$

$3 {x}^{2} \textcolor{b l u e}{+} \textcolor{b l u e}{9 x} = \textcolor{red}{\cancel{\textcolor{b l a c k}{\textcolor{b l a c k}{-} 9 x \textcolor{b l u e}{+} \textcolor{b l u e}{9 x}}}}$

$3 {x}^{2} + 9 x = 0$

Now, divide by the common factor $3$:

$\textcolor{b l u e}{\frac{\textcolor{b l a c k}{3 {x}^{2} + 9 x}}{3}} = \textcolor{b l u e}{\frac{\textcolor{b l a c k}{0}}{3}}$

$\textcolor{b l u e}{\frac{\textcolor{b l a c k}{\textcolor{red}{\cancel{\textcolor{b l a c k}{3}}} {x}^{2} + {\textcolor{red}{\cancel{\textcolor{b l a c k}{9}}}}^{3} x}}{\textcolor{red}{\cancel{\textcolor{b l u e}{3}}}}} = \textcolor{b l u e}{\frac{\textcolor{b l a c k}{0}}{3}}$

${x}^{2} + 3 x = \textcolor{b l u e}{\frac{\textcolor{b l a c k}{0}}{3}}$

${x}^{2} + 3 x = 0$

Now, use the distributive property backward to factor out the $x$ term:

$\textcolor{red}{x} \cdot \textcolor{b l u e}{x} + \textcolor{red}{x} \cdot \textcolor{b l u e}{3} = 0$

$\left(\textcolor{red}{x}\right) \left(\textcolor{b l u e}{x + 3}\right) = 0$

Now, set each factor equal to $0$ and solve for $x$ in each one:

color(white){color(black)( (color(red)x=0,qquadcolor(blue)(x+3)=0), (,qquadcolor(blue)x=-3):}

These are the two solutions: $x = 0$ and $x = - 3$. Hope this helped!