Solve 1/x-2x=5

$$\frac { 1 } { x } - 2 x = 5$$

Solve for x

$$x=\frac{\sqrt{33}-5}{4}\approx 0.186140662$$

$$x=\frac{-\sqrt{33}-5}{4}\approx -2.686140662$$

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.

$$1-2xx=5x$$

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

$$1-2x^{2}=5x$$

Subtract 5x from both sides.

$$1-2x^{2}-5x=0$$

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.

$$-2x^{2}-5x+1=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -2 for a, -5 for b, and 1 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\left(-2\right)}}{2\left(-2\right)}$$

Square -5.

$$x=\frac{-\left(-5\right)±\sqrt{25-4\left(-2\right)}}{2\left(-2\right)}$$

Multiply -4 times -2.

$$x=\frac{-\left(-5\right)±\sqrt{25+8}}{2\left(-2\right)}$$

Add 25 to 8.

$$x=\frac{-\left(-5\right)±\sqrt{33}}{2\left(-2\right)}$$

The opposite of -5 is 5.

$$x=\frac{5±\sqrt{33}}{2\left(-2\right)}$$

Multiply 2 times -2.

$$x=\frac{5±\sqrt{33}}{-4}$$

Now solve the equation x=\frac{5±\sqrt{33}}{-4} when ± is plus. Add 5 to \sqrt{33}.

$$x=\frac{\sqrt{33}+5}{-4}$$

Divide 5+\sqrt{33} by -4.

$$x=\frac{-\sqrt{33}-5}{4}$$

Now solve the equation x=\frac{5±\sqrt{33}}{-4} when ± is minus. Subtract \sqrt{33} from 5.

$$x=\frac{5-\sqrt{33}}{-4}$$

Divide 5-\sqrt{33} by -4.

$$x=\frac{\sqrt{33}-5}{4}$$

The equation is now solved.

$$x=\frac{-\sqrt{33}-5}{4} x=\frac{\sqrt{33}-5}{4}$$

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.

$$1-2xx=5x$$

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

$$1-2x^{2}=5x$$

Subtract 5x from both sides.

$$1-2x^{2}-5x=0$$

Subtract 1 from both sides. Anything subtracted from zero gives its negation.

$$-2x^{2}-5x=-1$$

Divide both sides by -2.

$$\frac{-2x^{2}-5x}{-2}=-\frac{1}{-2}$$

Dividing by -2 undoes the multiplication by -2.

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

Divide -5 by -2.

$$x^{2}+\frac{5}{2}x=-\frac{1}{-2}$$

Divide -1 by -2.

$$x^{2}+\frac{5}{2}x=\frac{1}{2}$$

Divide \frac{5}{2}, the coefficient of the x term, by 2 to get \frac{5}{4}. Then add the square of \frac{5}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{5}{2}x+\left(\frac{5}{4}\right)^{2}=\frac{1}{2}+\left(\frac{5}{4}\right)^{2}$$

Square \frac{5}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{5}{2}x+\frac{25}{16}=\frac{1}{2}+\frac{25}{16}$$

Add \frac{1}{2} to \frac{25}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{5}{2}x+\frac{25}{16}=\frac{33}{16}$$

Factor x^{2}+\frac{5}{2}x+\frac{25}{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}.

$$\left(x+\frac{5}{4}\right)^{2}=\frac{33}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{5}{4}\right)^{2}}=\sqrt{\frac{33}{16}}$$

Simplify.

$$x+\frac{5}{4}=\frac{\sqrt{33}}{4} x+\frac{5}{4}=-\frac{\sqrt{33}}{4}$$

Subtract \frac{5}{4} from both sides of the equation.

$$x=\frac{\sqrt{33}-5}{4} x=\frac{-\sqrt{33}-5}{4}$$