Solve 1/4x(2/3x-9)=0

$$\frac { 1 } { 4 } x ( \frac { 2 } { 3 } x - 9 ) = 0$$

Solve for x

$$x = \frac{27}{2} = 13\frac{1}{2} = 13.5$$

$$x=0$$

Steps Using Factoring

Use the distributive property to multiply \frac{1}{4}x by \frac{2}{3}x-9.

$$\frac{1}{4}x\times \frac{2}{3}x+\frac{1}{4}x\left(-9\right)=0$$

Multiply x and x to get x^{2}.

$$\frac{1}{4}x^{2}\times \frac{2}{3}+\frac{1}{4}x\left(-9\right)=0$$

Multiply \frac{1}{4} times \frac{2}{3} by multiplying numerator times numerator and denominator times denominator.

$$\frac{1\times 2}{4\times 3}x^{2}+\frac{1}{4}x\left(-9\right)=0$$

Do the multiplications in the fraction \frac{1\times 2}{4\times 3}.

$$\frac{2}{12}x^{2}+\frac{1}{4}x\left(-9\right)=0$$

Reduce the fraction \frac{2}{12} to lowest terms by extracting and canceling out 2.

$$\frac{1}{6}x^{2}+\frac{1}{4}x\left(-9\right)=0$$

Multiply \frac{1}{4} and -9 to get \frac{-9}{4}.

$$\frac{1}{6}x^{2}+\frac{-9}{4}x=0$$

Fraction \frac{-9}{4} can be rewritten as -\frac{9}{4} by extracting the negative sign.

$$\frac{1}{6}x^{2}-\frac{9}{4}x=0$$

Factor out x.

$$x\left(\frac{1}{6}x-\frac{9}{4}\right)=0$$

To find equation solutions, solve x=0 and \frac{x}{6}-\frac{9}{4}=0.

$$x=0 x=\frac{27}{2}$$

Steps Using the Quadratic Formula

Use the distributive property to multiply \frac{1}{4}x by \frac{2}{3}x-9.

$$\frac{1}{4}x\times \frac{2}{3}x+\frac{1}{4}x\left(-9\right)=0$$

Multiply x and x to get x^{2}.

$$\frac{1}{4}x^{2}\times \frac{2}{3}+\frac{1}{4}x\left(-9\right)=0$$

Multiply \frac{1}{4} times \frac{2}{3} by multiplying numerator times numerator and denominator times denominator.

$$\frac{1\times 2}{4\times 3}x^{2}+\frac{1}{4}x\left(-9\right)=0$$

Do the multiplications in the fraction \frac{1\times 2}{4\times 3}.

$$\frac{2}{12}x^{2}+\frac{1}{4}x\left(-9\right)=0$$

Reduce the fraction \frac{2}{12} to lowest terms by extracting and canceling out 2.

$$\frac{1}{6}x^{2}+\frac{1}{4}x\left(-9\right)=0$$

Multiply \frac{1}{4} and -9 to get \frac{-9}{4}.

$$\frac{1}{6}x^{2}+\frac{-9}{4}x=0$$

Fraction \frac{-9}{4} can be rewritten as -\frac{9}{4} by extracting the negative sign.

$$\frac{1}{6}x^{2}-\frac{9}{4}x=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute \frac{1}{6} for a, -\frac{9}{4} for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-\frac{9}{4}\right)±\sqrt{\left(-\frac{9}{4}\right)^{2}}}{2\times \frac{1}{6}}$$

Take the square root of \left(-\frac{9}{4}\right)^{2}.

$$x=\frac{-\left(-\frac{9}{4}\right)±\frac{9}{4}}{2\times \frac{1}{6}}$$

The opposite of -\frac{9}{4} is \frac{9}{4}.

$$x=\frac{\frac{9}{4}±\frac{9}{4}}{2\times \frac{1}{6}}$$

Multiply 2 times \frac{1}{6}.

$$x=\frac{\frac{9}{4}±\frac{9}{4}}{\frac{1}{3}}$$

Now solve the equation x=\frac{\frac{9}{4}±\frac{9}{4}}{\frac{1}{3}} when ± is plus. Add \frac{9}{4} to \frac{9}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=\frac{\frac{9}{2}}{\frac{1}{3}}$$

Divide \frac{9}{2} by \frac{1}{3} by multiplying \frac{9}{2} by the reciprocal of \frac{1}{3}.

$$x=\frac{27}{2}$$

Now solve the equation x=\frac{\frac{9}{4}±\frac{9}{4}}{\frac{1}{3}} when ± is minus. Subtract \frac{9}{4} from \frac{9}{4} by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$x=\frac{0}{\frac{1}{3}}$$

Divide 0 by \frac{1}{3} by multiplying 0 by the reciprocal of \frac{1}{3}.

$$x=0$$

The equation is now solved.

$$x=\frac{27}{2} x=0$$

Steps for Completing the Square

Use the distributive property to multiply \frac{1}{4}x by \frac{2}{3}x-9.

$$\frac{1}{4}x\times \frac{2}{3}x+\frac{1}{4}x\left(-9\right)=0$$

Multiply x and x to get x^{2}.

$$\frac{1}{4}x^{2}\times \frac{2}{3}+\frac{1}{4}x\left(-9\right)=0$$

Multiply \frac{1}{4} times \frac{2}{3} by multiplying numerator times numerator and denominator times denominator.

$$\frac{1\times 2}{4\times 3}x^{2}+\frac{1}{4}x\left(-9\right)=0$$

Do the multiplications in the fraction \frac{1\times 2}{4\times 3}.

$$\frac{2}{12}x^{2}+\frac{1}{4}x\left(-9\right)=0$$

Reduce the fraction \frac{2}{12} to lowest terms by extracting and canceling out 2.

$$\frac{1}{6}x^{2}+\frac{1}{4}x\left(-9\right)=0$$

Multiply \frac{1}{4} and -9 to get \frac{-9}{4}.

$$\frac{1}{6}x^{2}+\frac{-9}{4}x=0$$

Fraction \frac{-9}{4} can be rewritten as -\frac{9}{4} by extracting the negative sign.

$$\frac{1}{6}x^{2}-\frac{9}{4}x=0$$

Multiply both sides by 6.

$$\frac{\frac{1}{6}x^{2}-\frac{9}{4}x}{\frac{1}{6}}=\frac{0}{\frac{1}{6}}$$

Dividing by \frac{1}{6} undoes the multiplication by \frac{1}{6}.

$$x^{2}+\left(-\frac{\frac{9}{4}}{\frac{1}{6}}\right)x=\frac{0}{\frac{1}{6}}$$

Divide -\frac{9}{4} by \frac{1}{6} by multiplying -\frac{9}{4} by the reciprocal of \frac{1}{6}.

$$x^{2}-\frac{27}{2}x=\frac{0}{\frac{1}{6}}$$

Divide 0 by \frac{1}{6} by multiplying 0 by the reciprocal of \frac{1}{6}.

$$x^{2}-\frac{27}{2}x=0$$

Divide -\frac{27}{2}, the coefficient of the x term, by 2 to get -\frac{27}{4}. Then add the square of -\frac{27}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{27}{2}x+\left(-\frac{27}{4}\right)^{2}=\left(-\frac{27}{4}\right)^{2}$$

Square -\frac{27}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{27}{2}x+\frac{729}{16}=\frac{729}{16}$$

Factor x^{2}-\frac{27}{2}x+\frac{729}{16}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{27}{4}\right)^{2}=\frac{729}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{27}{4}\right)^{2}}=\sqrt{\frac{729}{16}}$$

Simplify.

$$x-\frac{27}{4}=\frac{27}{4} x-\frac{27}{4}=-\frac{27}{4}$$

Add \frac{27}{4} to both sides of the equation.

$$x=\frac{27}{2} x=0$$