Solve 9x^2=4x

$$9 x ^ { 2 } = 4 x$$

Solve for x

$$x=\frac{4}{9}\approx 0.444444444$$

$$x=0$$

Steps Using Factoring

Subtract 4x from both sides.

$$9x^{2}-4x=0$$

Factor out x.

$$x\left(9x-4\right)=0$$

To find equation solutions, solve x=0 and 9x-4=0.

$$x=0 x=\frac{4}{9}$$

Steps Using the Quadratic Formula

Subtract 4x from both sides.

$$9x^{2}-4x=0$$

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

$$x=\frac{-\left(-4\right)±\sqrt{\left(-4\right)^{2}}}{2\times 9}$$

Take the square root of \left(-4\right)^{2}.

$$x=\frac{-\left(-4\right)±4}{2\times 9}$$

The opposite of -4 is 4.

$$x=\frac{4±4}{2\times 9}$$

Multiply 2 times 9.

$$x=\frac{4±4}{18}$$

Now solve the equation x=\frac{4±4}{18} when ± is plus. Add 4 to 4.

$$x=\frac{8}{18}$$

Reduce the fraction \frac{8}{18} to lowest terms by extracting and canceling out 2.

$$x=\frac{4}{9}$$

Now solve the equation x=\frac{4±4}{18} when ± is minus. Subtract 4 from 4.

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

Divide 0 by 18.

$$x=0$$

The equation is now solved.

$$x=\frac{4}{9} x=0$$

Steps for Completing the Square

Subtract 4x from both sides.

$$9x^{2}-4x=0$$

Divide both sides by 9.

$$\frac{9x^{2}-4x}{9}=\frac{0}{9}$$

Dividing by 9 undoes the multiplication by 9.

$$x^{2}-\frac{4}{9}x=\frac{0}{9}$$

Divide 0 by 9.

$$x^{2}-\frac{4}{9}x=0$$

Divide -\frac{4}{9}, the coefficient of the x term, by 2 to get -\frac{2}{9}. Then add the square of -\frac{2}{9} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{4}{9}x+\left(-\frac{2}{9}\right)^{2}=\left(-\frac{2}{9}\right)^{2}$$

Square -\frac{2}{9} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{4}{9}x+\frac{4}{81}=\frac{4}{81}$$

Factor x^{2}-\frac{4}{9}x+\frac{4}{81}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{2}{9}\right)^{2}=\frac{4}{81}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{2}{9}\right)^{2}}=\sqrt{\frac{4}{81}}$$

Simplify.

$$x-\frac{2}{9}=\frac{2}{9} x-\frac{2}{9}=-\frac{2}{9}$$

Add \frac{2}{9} to both sides of the equation.

$$x=\frac{4}{9} x=0$$