Solve 8=(x-3/x)^2

$$8 = ( \frac { x - 3 } { x } ) ^ { 2 }$$

Solve for x

$$x=\frac{6\sqrt{2}-3}{7}\approx 0.783611625$$

$$x=\frac{-6\sqrt{2}-3}{7}\approx -1.640754482$$

Steps Using the Quadratic Formula

To raise \frac{x-3}{x} to a power, raise both numerator and denominator to the power and then divide.

$$8=\frac{\left(x-3\right)^{2}}{x^{2}}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-3\right)^{2}.

$$8=\frac{x^{2}-6x+9}{x^{2}}$$

Swap sides so that all variable terms are on the left hand side.

$$\frac{x^{2}-6x+9}{x^{2}}=8$$

Subtract 8 from both sides.

$$\frac{x^{2}-6x+9}{x^{2}}-8=0$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 8 times \frac{x^{2}}{x^{2}}.

$$\frac{x^{2}-6x+9}{x^{2}}-\frac{8x^{2}}{x^{2}}=0$$

Since \frac{x^{2}-6x+9}{x^{2}} and \frac{8x^{2}}{x^{2}} have the same denominator, subtract them by subtracting their numerators.

$$\frac{x^{2}-6x+9-8x^{2}}{x^{2}}=0$$

Combine like terms in x^{2}-6x+9-8x^{2}.

$$\frac{-7x^{2}-6x+9}{x^{2}}=0$$

Variable x cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by x^{2}.

$$-7x^{2}-6x+9=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -7 for a, -6 for b, and 9 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-6\right)±\sqrt{\left(-6\right)^{2}-4\left(-7\right)\times 9}}{2\left(-7\right)}$$

Square -6.

$$x=\frac{-\left(-6\right)±\sqrt{36-4\left(-7\right)\times 9}}{2\left(-7\right)}$$

Multiply -4 times -7.

$$x=\frac{-\left(-6\right)±\sqrt{36+28\times 9}}{2\left(-7\right)}$$

Multiply 28 times 9.

$$x=\frac{-\left(-6\right)±\sqrt{36+252}}{2\left(-7\right)}$$

Add 36 to 252.

$$x=\frac{-\left(-6\right)±\sqrt{288}}{2\left(-7\right)}$$

Take the square root of 288.

$$x=\frac{-\left(-6\right)±12\sqrt{2}}{2\left(-7\right)}$$

The opposite of -6 is 6.

$$x=\frac{6±12\sqrt{2}}{2\left(-7\right)}$$

Multiply 2 times -7.

$$x=\frac{6±12\sqrt{2}}{-14}$$

Now solve the equation x=\frac{6±12\sqrt{2}}{-14} when ± is plus. Add 6 to 12\sqrt{2}.

$$x=\frac{12\sqrt{2}+6}{-14}$$

Divide 12\sqrt{2}+6 by -14.

$$x=\frac{-6\sqrt{2}-3}{7}$$

Now solve the equation x=\frac{6±12\sqrt{2}}{-14} when ± is minus. Subtract 12\sqrt{2} from 6.

$$x=\frac{6-12\sqrt{2}}{-14}$$

Divide 6-12\sqrt{2} by -14.

$$x=\frac{6\sqrt{2}-3}{7}$$

The equation is now solved.

$$x=\frac{-6\sqrt{2}-3}{7} x=\frac{6\sqrt{2}-3}{7}$$

Steps for Completing the Square

To raise \frac{x-3}{x} to a power, raise both numerator and denominator to the power and then divide.

$$8=\frac{\left(x-3\right)^{2}}{x^{2}}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-3\right)^{2}.

$$8=\frac{x^{2}-6x+9}{x^{2}}$$

Swap sides so that all variable terms are on the left hand side.

$$\frac{x^{2}-6x+9}{x^{2}}=8$$

Variable x cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by x^{2}.

$$x^{2}-6x+9=8x^{2}$$

Subtract 8x^{2} from both sides.

$$x^{2}-6x+9-8x^{2}=0$$

Combine x^{2} and -8x^{2} to get -7x^{2}.

$$-7x^{2}-6x+9=0$$

Subtract 9 from both sides. Anything subtracted from zero gives its negation.

$$-7x^{2}-6x=-9$$

Divide both sides by -7.

$$\frac{-7x^{2}-6x}{-7}=-\frac{9}{-7}$$

Dividing by -7 undoes the multiplication by -7.

$$x^{2}+\left(-\frac{6}{-7}\right)x=-\frac{9}{-7}$$

Divide -6 by -7.

$$x^{2}+\frac{6}{7}x=-\frac{9}{-7}$$

Divide -9 by -7.

$$x^{2}+\frac{6}{7}x=\frac{9}{7}$$

Divide \frac{6}{7}, the coefficient of the x term, by 2 to get \frac{3}{7}. Then add the square of \frac{3}{7} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{6}{7}x+\left(\frac{3}{7}\right)^{2}=\frac{9}{7}+\left(\frac{3}{7}\right)^{2}$$

Square \frac{3}{7} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{6}{7}x+\frac{9}{49}=\frac{9}{7}+\frac{9}{49}$$

Add \frac{9}{7} to \frac{9}{49} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{6}{7}x+\frac{9}{49}=\frac{72}{49}$$

Factor x^{2}+\frac{6}{7}x+\frac{9}{49}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{3}{7}\right)^{2}=\frac{72}{49}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{3}{7}\right)^{2}}=\sqrt{\frac{72}{49}}$$

Simplify.

$$x+\frac{3}{7}=\frac{6\sqrt{2}}{7} x+\frac{3}{7}=-\frac{6\sqrt{2}}{7}$$

Subtract \frac{3}{7} from both sides of the equation.

$$x=\frac{6\sqrt{2}-3}{7} x=\frac{-6\sqrt{2}-3}{7}$$