Solve x^2=1/3x+2

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

Solve for x

$$x = \frac{\sqrt{73} + 1}{6} \approx 1.590667291$$

$$x=\frac{1-\sqrt{73}}{6}\approx -1.257333958$$

Steps Using the Quadratic Formula

Subtract \frac{1}{3}x from both sides.

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

Subtract 2 from both sides.

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

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

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

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

$$x=\frac{-\left(-\frac{1}{3}\right)±\sqrt{\frac{1}{9}-4\left(-2\right)}}{2}$$

Multiply -4 times -2.

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

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

$$x=\frac{-\left(-\frac{1}{3}\right)±\sqrt{\frac{73}{9}}}{2}$$

Take the square root of \frac{73}{9}.

$$x=\frac{-\left(-\frac{1}{3}\right)±\frac{\sqrt{73}}{3}}{2}$$

The opposite of -\frac{1}{3} is \frac{1}{3}.

$$x=\frac{\frac{1}{3}±\frac{\sqrt{73}}{3}}{2}$$

Now solve the equation x=\frac{\frac{1}{3}±\frac{\sqrt{73}}{3}}{2} when ± is plus. Add \frac{1}{3} to \frac{\sqrt{73}}{3}.

$$x=\frac{\sqrt{73}+1}{2\times 3}$$

Divide \frac{1+\sqrt{73}}{3} by 2.

$$x=\frac{\sqrt{73}+1}{6}$$

Now solve the equation x=\frac{\frac{1}{3}±\frac{\sqrt{73}}{3}}{2} when ± is minus. Subtract \frac{\sqrt{73}}{3} from \frac{1}{3}.

$$x=\frac{1-\sqrt{73}}{2\times 3}$$

Divide \frac{1-\sqrt{73}}{3} by 2.

$$x=\frac{1-\sqrt{73}}{6}$$

The equation is now solved.

$$x=\frac{\sqrt{73}+1}{6} x=\frac{1-\sqrt{73}}{6}$$

Steps for Completing the Square

Subtract \frac{1}{3}x from both sides.

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

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.

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

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

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

Add 2 to \frac{1}{36}.

$$x^{2}-\frac{1}{3}x+\frac{1}{36}=\frac{73}{36}$$

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}.

$$\left(x-\frac{1}{6}\right)^{2}=\frac{73}{36}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{6}\right)^{2}}=\sqrt{\frac{73}{36}}$$

Simplify.

$$x-\frac{1}{6}=\frac{\sqrt{73}}{6} x-\frac{1}{6}=-\frac{\sqrt{73}}{6}$$

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

$$x=\frac{\sqrt{73}+1}{6} x=\frac{1-\sqrt{73}}{6}$$