Solve x-1/x=10

$$x- \frac{ 1 }{ x } =10$$

Solve for x

$$x=\sqrt{26}+5\approx 10.099019514$$

$$x=5-\sqrt{26}\approx -0.099019514$$

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.

$$xx-1=10x$$

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

$$x^{2}-1=10x$$

Subtract 10x from both sides.

$$x^{2}-1-10x=0$$

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.

$$x^{2}-10x-1=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -10 for b, and -1 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-10\right)±\sqrt{\left(-10\right)^{2}-4\left(-1\right)}}{2}$$

Square -10.

$$x=\frac{-\left(-10\right)±\sqrt{100-4\left(-1\right)}}{2}$$

Multiply -4 times -1.

$$x=\frac{-\left(-10\right)±\sqrt{100+4}}{2}$$

Add 100 to 4.

$$x=\frac{-\left(-10\right)±\sqrt{104}}{2}$$

Take the square root of 104.

$$x=\frac{-\left(-10\right)±2\sqrt{26}}{2}$$

The opposite of -10 is 10.

$$x=\frac{10±2\sqrt{26}}{2}$$

Now solve the equation x=\frac{10±2\sqrt{26}}{2} when ± is plus. Add 10 to 2\sqrt{26}.

$$x=\frac{2\sqrt{26}+10}{2}$$

Divide 10+2\sqrt{26} by 2.

$$x=\sqrt{26}+5$$

Now solve the equation x=\frac{10±2\sqrt{26}}{2} when ± is minus. Subtract 2\sqrt{26} from 10.

$$x=\frac{10-2\sqrt{26}}{2}$$

Divide 10-2\sqrt{26} by 2.

$$x=5-\sqrt{26}$$

The equation is now solved.

$$x=\sqrt{26}+5 x=5-\sqrt{26}$$

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.

$$xx-1=10x$$

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

$$x^{2}-1=10x$$

Subtract 10x from both sides.

$$x^{2}-1-10x=0$$

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

$$x^{2}-10x=1$$

Divide -10, the coefficient of the x term, by 2 to get -5. Then add the square of -5 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-10x+\left(-5\right)^{2}=1+\left(-5\right)^{2}$$

Square -5.

$$x^{2}-10x+25=1+25$$

Add 1 to 25.

$$x^{2}-10x+25=26$$

Factor x^{2}-10x+25. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-5\right)^{2}=26$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-5\right)^{2}}=\sqrt{26}$$

Simplify.

$$x-5=\sqrt{26} x-5=-\sqrt{26}$$

Add 5 to both sides of the equation.

$$x=\sqrt{26}+5 x=5-\sqrt{26}$$