Solve 9(x^2)=3x

$$9 ( x ^ { 2 } ) = 3 x$$

Solve for x

$$x=\frac{1}{3}\approx 0.333333333$$

$$x=0$$

Steps Using Factoring

Subtract 3x from both sides.

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

Factor out x.

$$x\left(9x-3\right)=0$$

To find equation solutions, solve x=0 and 9x-3=0.

$$x=0 x=\frac{1}{3}$$

Steps Using the Quadratic Formula

Subtract 3x from both sides.

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

This equation is in standard form: ax^{2}+bx+c=0. Substitute 9 for a, -3 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}}}{2\times 9}$$

Take the square root of \left(-3\right)^{2}.

$$x=\frac{-\left(-3\right)±3}{2\times 9}$$

The opposite of -3 is 3.

$$x=\frac{3±3}{2\times 9}$$

Multiply 2 times 9.

$$x=\frac{3±3}{18}$$

Now solve the equation x=\frac{3±3}{18} when ± is plus. Add 3 to 3.

$$x=\frac{6}{18}$$

Reduce the fraction \frac{6}{18} to lowest terms by extracting and canceling out 6.

$$x=\frac{1}{3}$$

Now solve the equation x=\frac{3±3}{18} when ± is minus. Subtract 3 from 3.

$$x=\frac{0}{18}$$

Divide 0 by 18.

$$x=0$$

The equation is now solved.

$$x=\frac{1}{3} x=0$$

Steps for Completing the Square

Subtract 3x from both sides.

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

Divide both sides by 9.

$$\frac{9x^{2}-3x}{9}=\frac{0}{9}$$

Dividing by 9 undoes the multiplication by 9.

$$x^{2}+\left(-\frac{3}{9}\right)x=\frac{0}{9}$$

Reduce the fraction \frac{-3}{9} to lowest terms by extracting and canceling out 3.

$$x^{2}-\frac{1}{3}x=\frac{0}{9}$$

Divide 0 by 9.

$$x^{2}-\frac{1}{3}x=0$$

Divide -\frac{1}{3}, the coefficient of the x term, by 2 to get -\frac{1}{6}. Then add the square of -\frac{1}{6} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{1}{3}x+\left(-\frac{1}{6}\right)^{2}=\left(-\frac{1}{6}\right)^{2}$$

Square -\frac{1}{6} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{1}{3}x+\frac{1}{36}=\frac{1}{36}$$

Factor x^{2}-\frac{1}{3}x+\frac{1}{36}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{1}{6}\right)^{2}=\frac{1}{36}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{6}\right)^{2}}=\sqrt{\frac{1}{36}}$$

Simplify.

$$x-\frac{1}{6}=\frac{1}{6} x-\frac{1}{6}=-\frac{1}{6}$$

Add \frac{1}{6} to both sides of the equation.

$$x=\frac{1}{3} x=0$$