Solve x-1/x=11

$$x- \frac{ 1 }{ x } =11$$

Solve for x

$$x = \frac{5 \sqrt{5} + 11}{2} \approx 11.090169944$$

$$x=\frac{11-5\sqrt{5}}{2}\approx -0.090169944$$

Steps Using the Quadratic Formula

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

$$xx-1=11x$$

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

$$x^{2}-1=11x$$

Subtract 11x from both sides.

$$x^{2}-1-11x=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$x^{2}-11x-1=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -11 for b, and -1 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-11\right)±\sqrt{\left(-11\right)^{2}-4\left(-1\right)}}{2}$$

Square -11.

$$x=\frac{-\left(-11\right)±\sqrt{121-4\left(-1\right)}}{2}$$

Multiply -4 times -1.

$$x=\frac{-\left(-11\right)±\sqrt{121+4}}{2}$$

Add 121 to 4.

$$x=\frac{-\left(-11\right)±\sqrt{125}}{2}$$

Take the square root of 125.

$$x=\frac{-\left(-11\right)±5\sqrt{5}}{2}$$

The opposite of -11 is 11.

$$x=\frac{11±5\sqrt{5}}{2}$$

Now solve the equation x=\frac{11±5\sqrt{5}}{2} when ± is plus. Add 11 to 5\sqrt{5}.

$$x=\frac{5\sqrt{5}+11}{2}$$

Now solve the equation x=\frac{11±5\sqrt{5}}{2} when ± is minus. Subtract 5\sqrt{5} from 11.

$$x=\frac{11-5\sqrt{5}}{2}$$

The equation is now solved.

$$x=\frac{5\sqrt{5}+11}{2} x=\frac{11-5\sqrt{5}}{2}$$

Steps for Completing the Square

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

$$xx-1=11x$$

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

$$x^{2}-1=11x$$

Subtract 11x from both sides.

$$x^{2}-1-11x=0$$

Add 1 to both sides. Anything plus zero gives itself.

$$x^{2}-11x=1$$

Divide -11, the coefficient of the x term, by 2 to get -\frac{11}{2}. Then add the square of -\frac{11}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-11x+\left(-\frac{11}{2}\right)^{2}=1+\left(-\frac{11}{2}\right)^{2}$$

Square -\frac{11}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-11x+\frac{121}{4}=1+\frac{121}{4}$$

Add 1 to \frac{121}{4}.

$$x^{2}-11x+\frac{121}{4}=\frac{125}{4}$$

Factor x^{2}-11x+\frac{121}{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}.

$$\left(x-\frac{11}{2}\right)^{2}=\frac{125}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{11}{2}\right)^{2}}=\sqrt{\frac{125}{4}}$$

Simplify.

$$x-\frac{11}{2}=\frac{5\sqrt{5}}{2} x-\frac{11}{2}=-\frac{5\sqrt{5}}{2}$$

Add \frac{11}{2} to both sides of the equation.

$$x=\frac{5\sqrt{5}+11}{2} x=\frac{11-5\sqrt{5}}{2}$$