Solve 1/3x^2+2/3=x

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

Solve for x

$$x=1$$

$$x=2$$

Steps Using the Quadratic Formula

Subtract x from both sides.

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

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

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

This equation is in standard form: ax^{2}+bx+c=0. Substitute \frac{1}{3} for a, -1 for b, and \frac{2}{3} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-1\right)±\sqrt{1-4\times \frac{1}{3}\times \frac{2}{3}}}{2\times \frac{1}{3}}$$

Multiply -4 times \frac{1}{3}.

$$x=\frac{-\left(-1\right)±\sqrt{1-\frac{4}{3}\times \frac{2}{3}}}{2\times \frac{1}{3}}$$

Multiply -\frac{4}{3} times \frac{2}{3} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$x=\frac{-\left(-1\right)±\sqrt{1-\frac{8}{9}}}{2\times \frac{1}{3}}$$

Add 1 to -\frac{8}{9}.

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

Take the square root of \frac{1}{9}.

$$x=\frac{-\left(-1\right)±\frac{1}{3}}{2\times \frac{1}{3}}$$

The opposite of -1 is 1.

$$x=\frac{1±\frac{1}{3}}{2\times \frac{1}{3}}$$

Multiply 2 times \frac{1}{3}.

$$x=\frac{1±\frac{1}{3}}{\frac{2}{3}}$$

Now solve the equation x=\frac{1±\frac{1}{3}}{\frac{2}{3}} when ± is plus. Add 1 to \frac{1}{3}.

$$x=\frac{\frac{4}{3}}{\frac{2}{3}}$$

Divide \frac{4}{3} by \frac{2}{3} by multiplying \frac{4}{3} by the reciprocal of \frac{2}{3}.

$$x=2$$

Now solve the equation x=\frac{1±\frac{1}{3}}{\frac{2}{3}} when ± is minus. Subtract \frac{1}{3} from 1.

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

Divide \frac{2}{3} by \frac{2}{3} by multiplying \frac{2}{3} by the reciprocal of \frac{2}{3}.

$$x=1$$

The equation is now solved.

$$x=2 x=1$$

Steps for Completing the Square

Subtract x from both sides.

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

Subtract \frac{2}{3} from both sides. Anything subtracted from zero gives its negation.

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

Multiply both sides by 3.

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

Dividing by \frac{1}{3} undoes the multiplication by \frac{1}{3}.

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

Divide -1 by \frac{1}{3} by multiplying -1 by the reciprocal of \frac{1}{3}.

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

Divide -\frac{2}{3} by \frac{1}{3} by multiplying -\frac{2}{3} by the reciprocal of \frac{1}{3}.

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

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

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

Square -\frac{3}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-3x+\frac{9}{4}=-2+\frac{9}{4}$$

Add -2 to \frac{9}{4}.

$$x^{2}-3x+\frac{9}{4}=\frac{1}{4}$$

Factor x^{2}-3x+\frac{9}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{3}{2}\right)^{2}=\frac{1}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{3}{2}\right)^{2}}=\sqrt{\frac{1}{4}}$$

Simplify.

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

Add \frac{3}{2} to both sides of the equation.

$$x=2 x=1$$