Solve 3/4x^2-1/3x^2-5/3x=0

$$\frac{ 3 }{ 4 } { x }^{ 2 } - \frac{ 1 }{ 3 } { x }^{ 2 } - \frac{ 5 }{ 3 } x=0$$

Solve for x

$$x=4$$

$$x=0$$

Steps Using Factoring

Factor out x.

$$x\left(\frac{3}{4}x-\frac{1}{3}x-\frac{5}{3}\right)=0$$

To find equation solutions, solve x=0 and \frac{3x}{4}-\frac{x}{3}-\frac{5}{3}=0.

$$x=0 x=4$$

Steps Using the Quadratic Formula

Combine \frac{3}{4}x^{2} and -\frac{1}{3}x^{2} to get \frac{5}{12}x^{2}.

$$\frac{5}{12}x^{2}-\frac{5}{3}x=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute \frac{5}{12} for a, -\frac{5}{3} for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-\frac{5}{3}\right)±\sqrt{\left(-\frac{5}{3}\right)^{2}}}{2\times \frac{5}{12}}$$

Take the square root of \left(-\frac{5}{3}\right)^{2}.

$$x=\frac{-\left(-\frac{5}{3}\right)±\frac{5}{3}}{2\times \frac{5}{12}}$$

The opposite of -\frac{5}{3} is \frac{5}{3}.

$$x=\frac{\frac{5}{3}±\frac{5}{3}}{2\times \frac{5}{12}}$$

Multiply 2 times \frac{5}{12}.

$$x=\frac{\frac{5}{3}±\frac{5}{3}}{\frac{5}{6}}$$

Now solve the equation x=\frac{\frac{5}{3}±\frac{5}{3}}{\frac{5}{6}} when ± is plus. Add \frac{5}{3} to \frac{5}{3} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=\frac{\frac{10}{3}}{\frac{5}{6}}$$

Divide \frac{10}{3} by \frac{5}{6} by multiplying \frac{10}{3} by the reciprocal of \frac{5}{6}.

$$x=4$$

Now solve the equation x=\frac{\frac{5}{3}±\frac{5}{3}}{\frac{5}{6}} when ± is minus. Subtract \frac{5}{3} from \frac{5}{3} by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$x=\frac{0}{\frac{5}{6}}$$

Divide 0 by \frac{5}{6} by multiplying 0 by the reciprocal of \frac{5}{6}.

$$x=0$$

The equation is now solved.

$$x=4 x=0$$

Steps for Completing the Square

Combine \frac{3}{4}x^{2} and -\frac{1}{3}x^{2} to get \frac{5}{12}x^{2}.

$$\frac{5}{12}x^{2}-\frac{5}{3}x=0$$

Divide both sides of the equation by \frac{5}{12}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$\frac{\frac{5}{12}x^{2}-\frac{5}{3}x}{\frac{5}{12}}=\frac{0}{\frac{5}{12}}$$

Dividing by \frac{5}{12} undoes the multiplication by \frac{5}{12}.

$$x^{2}+\left(-\frac{\frac{5}{3}}{\frac{5}{12}}\right)x=\frac{0}{\frac{5}{12}}$$

Divide -\frac{5}{3} by \frac{5}{12} by multiplying -\frac{5}{3} by the reciprocal of \frac{5}{12}.

$$x^{2}-4x=\frac{0}{\frac{5}{12}}$$

Divide 0 by \frac{5}{12} by multiplying 0 by the reciprocal of \frac{5}{12}.

$$x^{2}-4x=0$$

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

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

Square -2.

$$x^{2}-4x+4=4$$

Factor x^{2}-4x+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}.

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

Take the square root of both sides of the equation.

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

Simplify.

$$x-2=2 x-2=-2$$

Add 2 to both sides of the equation.

$$x=4 x=0$$