Solve (x+2)/frac{6{x}}=8

$$\frac { ( x + 2 ) } { \frac { 6 } { x } } = 8$$

Solve for x

$$x=-8$$

$$x=6$$

Steps Using the Quadratic Formula

Variable x cannot be equal to 0 since division by zero is not defined. Divide x+2 by \frac{6}{x} by multiplying x+2 by the reciprocal of \frac{6}{x}.

$$\frac{\left(x+2\right)x}{6}=8$$

Use the distributive property to multiply x+2 by x.

$$\frac{x^{2}+2x}{6}=8$$

Divide each term of x^{2}+2x by 6 to get \frac{1}{6}x^{2}+\frac{1}{3}x.

$$\frac{1}{6}x^{2}+\frac{1}{3}x=8$$

Subtract 8 from both sides.

$$\frac{1}{6}x^{2}+\frac{1}{3}x-8=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute \frac{1}{6} for a, \frac{1}{3} for b, and -8 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\frac{1}{3}±\sqrt{\left(\frac{1}{3}\right)^{2}-4\times \frac{1}{6}\left(-8\right)}}{2\times \frac{1}{6}}$$

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

$$x=\frac{-\frac{1}{3}±\sqrt{\frac{1}{9}-4\times \frac{1}{6}\left(-8\right)}}{2\times \frac{1}{6}}$$

Multiply -4 times \frac{1}{6}.

$$x=\frac{-\frac{1}{3}±\sqrt{\frac{1}{9}-\frac{2}{3}\left(-8\right)}}{2\times \frac{1}{6}}$$

Multiply -\frac{2}{3} times -8.

$$x=\frac{-\frac{1}{3}±\sqrt{\frac{1}{9}+\frac{16}{3}}}{2\times \frac{1}{6}}$$

Add \frac{1}{9} to \frac{16}{3} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=\frac{-\frac{1}{3}±\sqrt{\frac{49}{9}}}{2\times \frac{1}{6}}$$

Take the square root of \frac{49}{9}.

$$x=\frac{-\frac{1}{3}±\frac{7}{3}}{2\times \frac{1}{6}}$$

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

$$x=\frac{-\frac{1}{3}±\frac{7}{3}}{\frac{1}{3}}$$

Now solve the equation x=\frac{-\frac{1}{3}±\frac{7}{3}}{\frac{1}{3}} when ± is plus. Add -\frac{1}{3} to \frac{7}{3} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

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

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

$$x=6$$

Now solve the equation x=\frac{-\frac{1}{3}±\frac{7}{3}}{\frac{1}{3}} when ± is minus. Subtract \frac{7}{3} from -\frac{1}{3} by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$x=-\frac{\frac{8}{3}}{\frac{1}{3}}$$

Divide -\frac{8}{3} by \frac{1}{3} by multiplying -\frac{8}{3} by the reciprocal of \frac{1}{3}.

$$x=-8$$

The equation is now solved.

$$x=6 x=-8$$

Steps for Completing the Square

Variable x cannot be equal to 0 since division by zero is not defined. Divide x+2 by \frac{6}{x} by multiplying x+2 by the reciprocal of \frac{6}{x}.

$$\frac{\left(x+2\right)x}{6}=8$$

Use the distributive property to multiply x+2 by x.

$$\frac{x^{2}+2x}{6}=8$$

Divide each term of x^{2}+2x by 6 to get \frac{1}{6}x^{2}+\frac{1}{3}x.

$$\frac{1}{6}x^{2}+\frac{1}{3}x=8$$

Multiply both sides by 6.

$$\frac{\frac{1}{6}x^{2}+\frac{1}{3}x}{\frac{1}{6}}=\frac{8}{\frac{1}{6}}$$

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

$$x^{2}+\frac{\frac{1}{3}}{\frac{1}{6}}x=\frac{8}{\frac{1}{6}}$$

Divide \frac{1}{3} by \frac{1}{6} by multiplying \frac{1}{3} by the reciprocal of \frac{1}{6}.

$$x^{2}+2x=\frac{8}{\frac{1}{6}}$$

Divide 8 by \frac{1}{6} by multiplying 8 by the reciprocal of \frac{1}{6}.

$$x^{2}+2x=48$$

Divide 2, the coefficient of the x term, by 2 to get 1. Then add the square of 1 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+2x+1^{2}=48+1^{2}$$

Square 1.

$$x^{2}+2x+1=48+1$$

Add 48 to 1.

$$x^{2}+2x+1=49$$

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

$$\left(x+1\right)^{2}=49$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+1\right)^{2}}=\sqrt{49}$$

Simplify.

$$x+1=7 x+1=-7$$

Subtract 1 from both sides of the equation.

$$x=6 x=-8$$