Solve 2x-5=5/x

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

Solve for x

$$x = \frac{\sqrt{65} + 5}{4} \approx 3.265564437$$

$$x=\frac{5-\sqrt{65}}{4}\approx -0.765564437$$

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.

$$2xx+x\left(-5\right)=5$$

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

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

Subtract 5 from both sides.

$$2x^{2}+x\left(-5\right)-5=0$$

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.

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

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

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

Square -5.

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

Multiply -4 times 2.

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

Multiply -8 times -5.

$$x=\frac{-\left(-5\right)±\sqrt{25+40}}{2\times 2}$$

Add 25 to 40.

$$x=\frac{-\left(-5\right)±\sqrt{65}}{2\times 2}$$

The opposite of -5 is 5.

$$x=\frac{5±\sqrt{65}}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{5±\sqrt{65}}{4}$$

Now solve the equation x=\frac{5±\sqrt{65}}{4} when ± is plus. Add 5 to \sqrt{65}.

$$x=\frac{\sqrt{65}+5}{4}$$

Now solve the equation x=\frac{5±\sqrt{65}}{4} when ± is minus. Subtract \sqrt{65} from 5.

$$x=\frac{5-\sqrt{65}}{4}$$

The equation is now solved.

$$x=\frac{\sqrt{65}+5}{4} x=\frac{5-\sqrt{65}}{4}$$

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.

$$2xx+x\left(-5\right)=5$$

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

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

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

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

Divide both sides by 2.

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

Dividing by 2 undoes the multiplication by 2.

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

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.

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

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

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

Add \frac{5}{2} to \frac{25}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{5}{2}x+\frac{25}{16}=\frac{65}{16}$$

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}.

$$\left(x-\frac{5}{4}\right)^{2}=\frac{65}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{5}{4}\right)^{2}}=\sqrt{\frac{65}{16}}$$

Simplify.

$$x-\frac{5}{4}=\frac{\sqrt{65}}{4} x-\frac{5}{4}=-\frac{\sqrt{65}}{4}$$

Add \frac{5}{4} to both sides of the equation.

$$x=\frac{\sqrt{65}+5}{4} x=\frac{5-\sqrt{65}}{4}$$