Solve 3A^2=1/3A

$$3 A ^ { 2 } = \frac { 1 } { 3 } A$$

Solve for A

$$A=\frac{1}{9}\approx 0.111111111$$

$$A=0$$

Steps Using Factoring

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

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

Factor out A.

$$A\left(3A-\frac{1}{3}\right)=0$$

To find equation solutions, solve A=0 and 3A-\frac{1}{3}=0.

$$A=0 A=\frac{1}{9}$$

Steps Using the Quadratic Formula

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

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

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

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

Take the square root of \left(-\frac{1}{3}\right)^{2}.

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

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

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

Multiply 2 times 3.

$$A=\frac{\frac{1}{3}±\frac{1}{3}}{6}$$

Now solve the equation A=\frac{\frac{1}{3}±\frac{1}{3}}{6} when ± is plus. Add \frac{1}{3} to \frac{1}{3} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$A=\frac{\frac{2}{3}}{6}$$

Divide \frac{2}{3} by 6.

$$A=\frac{1}{9}$$

Now solve the equation A=\frac{\frac{1}{3}±\frac{1}{3}}{6} when ± is minus. Subtract \frac{1}{3} from \frac{1}{3} by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

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

Divide 0 by 6.

$$A=0$$

The equation is now solved.

$$A=\frac{1}{9} A=0$$

Steps for Completing the Square

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

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

Divide both sides by 3.

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

Dividing by 3 undoes the multiplication by 3.

$$A^{2}+\left(-\frac{\frac{1}{3}}{3}\right)A=\frac{0}{3}$$

Divide -\frac{1}{3} by 3.

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

Divide 0 by 3.

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

Divide -\frac{1}{9}, the coefficient of the x term, by 2 to get -\frac{1}{18}. Then add the square of -\frac{1}{18} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$A^{2}-\frac{1}{9}A+\left(-\frac{1}{18}\right)^{2}=\left(-\frac{1}{18}\right)^{2}$$

Square -\frac{1}{18} by squaring both the numerator and the denominator of the fraction.

$$A^{2}-\frac{1}{9}A+\frac{1}{324}=\frac{1}{324}$$

Factor A^{2}-\frac{1}{9}A+\frac{1}{324}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(A-\frac{1}{18}\right)^{2}=\frac{1}{324}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(A-\frac{1}{18}\right)^{2}}=\sqrt{\frac{1}{324}}$$

Simplify.

$$A-\frac{1}{18}=\frac{1}{18} A-\frac{1}{18}=-\frac{1}{18}$$

Add \frac{1}{18} to both sides of the equation.

$$A=\frac{1}{9} A=0$$