Solve 3x^2=3/4x

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

Solve for x

$$x=\frac{1}{4}=0.25$$

$$x=0$$

Steps Using Factoring

Subtract \frac{3}{4}x from both sides.

$$3x^{2}-\frac{3}{4}x=0$$

Factor out x.

$$x\left(3x-\frac{3}{4}\right)=0$$

To find equation solutions, solve x=0 and 3x-\frac{3}{4}=0.

$$x=0 x=\frac{1}{4}$$

Steps Using the Quadratic Formula

Subtract \frac{3}{4}x from both sides.

$$3x^{2}-\frac{3}{4}x=0$$

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

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

Take the square root of \left(-\frac{3}{4}\right)^{2}.

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

The opposite of -\frac{3}{4} is \frac{3}{4}.

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

Multiply 2 times 3.

$$x=\frac{\frac{3}{4}±\frac{3}{4}}{6}$$

Now solve the equation x=\frac{\frac{3}{4}±\frac{3}{4}}{6} when ± is plus. Add \frac{3}{4} to \frac{3}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

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

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

$$x=\frac{1}{4}$$

Now solve the equation x=\frac{\frac{3}{4}±\frac{3}{4}}{6} when ± is minus. Subtract \frac{3}{4} from \frac{3}{4} by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

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

Divide 0 by 6.

$$x=0$$

The equation is now solved.

$$x=\frac{1}{4} x=0$$

Steps for Completing the Square

Subtract \frac{3}{4}x from both sides.

$$3x^{2}-\frac{3}{4}x=0$$

Divide both sides by 3.

$$\frac{3x^{2}-\frac{3}{4}x}{3}=\frac{0}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$x^{2}+\left(-\frac{\frac{3}{4}}{3}\right)x=\frac{0}{3}$$

Divide -\frac{3}{4} by 3.

$$x^{2}-\frac{1}{4}x=\frac{0}{3}$$

Divide 0 by 3.

$$x^{2}-\frac{1}{4}x=0$$

Divide -\frac{1}{4}, the coefficient of the x term, by 2 to get -\frac{1}{8}. Then add the square of -\frac{1}{8} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{1}{4}x+\left(-\frac{1}{8}\right)^{2}=\left(-\frac{1}{8}\right)^{2}$$

Square -\frac{1}{8} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{1}{4}x+\frac{1}{64}=\frac{1}{64}$$

Factor x^{2}-\frac{1}{4}x+\frac{1}{64}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{1}{8}\right)^{2}=\frac{1}{64}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{8}\right)^{2}}=\sqrt{\frac{1}{64}}$$

Simplify.

$$x-\frac{1}{8}=\frac{1}{8} x-\frac{1}{8}=-\frac{1}{8}$$

Add \frac{1}{8} to both sides of the equation.

$$x=\frac{1}{4} x=0$$