Solve x^2/2=x/3

$$\frac { x ^ { 2 } } { 2 } = \frac { x } { 3 }$$

Solve for x

$$x=\frac{2}{3}\approx 0.666666667$$

$$x=0$$

Steps Using Factoring

Multiply both sides of the equation by 6, the least common multiple of 2,3.

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

Subtract 2x from both sides.

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

Factor out x.

$$x\left(3x-2\right)=0$$

To find equation solutions, solve x=0 and 3x-2=0.

$$x=0 x=\frac{2}{3}$$

Steps Using the Quadratic Formula

Multiply both sides of the equation by 6, the least common multiple of 2,3.

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

Subtract 2x from both sides.

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

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

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

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

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

The opposite of -2 is 2.

$$x=\frac{2±2}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{2±2}{6}$$

Now solve the equation x=\frac{2±2}{6} when ± is plus. Add 2 to 2.

$$x=\frac{4}{6}$$

Reduce the fraction \frac{4}{6} to lowest terms by extracting and canceling out 2.

$$x=\frac{2}{3}$$

Now solve the equation x=\frac{2±2}{6} when ± is minus. Subtract 2 from 2.

$$x=\frac{0}{6}$$

Divide 0 by 6.

$$x=0$$

The equation is now solved.

$$x=\frac{2}{3} x=0$$

Steps for Completing the Square

Multiply both sides of the equation by 6, the least common multiple of 2,3.

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

Subtract 2x from both sides.

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

Divide both sides by 3.

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

Dividing by 3 undoes the multiplication by 3.

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

Divide 0 by 3.

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

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

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

Square -\frac{1}{3} by squaring both the numerator and the denominator of the fraction.

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

Factor x^{2}-\frac{2}{3}x+\frac{1}{9}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{1}{3}\right)^{2}=\frac{1}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{3}\right)^{2}}=\sqrt{\frac{1}{9}}$$

Simplify.

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

Add \frac{1}{3} to both sides of the equation.

$$x=\frac{2}{3} x=0$$