Solve x=25/x-1

$$x = \frac { 25 } { x } - 1$$

Solve for x

$$x = \frac{\sqrt{101} - 1}{2} \approx 4.524937811$$

$$x=\frac{-\sqrt{101}-1}{2}\approx -5.524937811$$

Steps Using the Quadratic Formula

To add or subtract expressions, expand them to make their denominators the same. Multiply 1 times \frac{x}{x}.

$$x=\frac{25}{x}-\frac{x}{x}$$

Since \frac{25}{x} and \frac{x}{x} have the same denominator, subtract them by subtracting their numerators.

$$x=\frac{25-x}{x}$$

Subtract \frac{25-x}{x} from both sides.

$$x-\frac{25-x}{x}=0$$

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

$$\frac{xx}{x}-\frac{25-x}{x}=0$$

Since \frac{xx}{x} and \frac{25-x}{x} have the same denominator, subtract them by subtracting their numerators.

$$\frac{xx-\left(25-x\right)}{x}=0$$

Do the multiplications in xx-\left(25-x\right).

$$\frac{x^{2}-25+x}{x}=0$$

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

$$x^{2}-25+x=0$$

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.

$$x^{2}+x-25=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 1 for b, and -25 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-1±\sqrt{1^{2}-4\left(-25\right)}}{2}$$

Square 1.

$$x=\frac{-1±\sqrt{1-4\left(-25\right)}}{2}$$

Multiply -4 times -25.

$$x=\frac{-1±\sqrt{1+100}}{2}$$

Add 1 to 100.

$$x=\frac{-1±\sqrt{101}}{2}$$

Now solve the equation x=\frac{-1±\sqrt{101}}{2} when ± is plus. Add -1 to \sqrt{101}.

$$x=\frac{\sqrt{101}-1}{2}$$

Now solve the equation x=\frac{-1±\sqrt{101}}{2} when ± is minus. Subtract \sqrt{101} from -1.

$$x=\frac{-\sqrt{101}-1}{2}$$

The equation is now solved.

$$x=\frac{\sqrt{101}-1}{2} x=\frac{-\sqrt{101}-1}{2}$$

Steps for Completing the Square

To add or subtract expressions, expand them to make their denominators the same. Multiply 1 times \frac{x}{x}.

$$x=\frac{25}{x}-\frac{x}{x}$$

Since \frac{25}{x} and \frac{x}{x} have the same denominator, subtract them by subtracting their numerators.

$$x=\frac{25-x}{x}$$

Subtract \frac{25-x}{x} from both sides.

$$x-\frac{25-x}{x}=0$$

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

$$\frac{xx}{x}-\frac{25-x}{x}=0$$

Since \frac{xx}{x} and \frac{25-x}{x} have the same denominator, subtract them by subtracting their numerators.

$$\frac{xx-\left(25-x\right)}{x}=0$$

Do the multiplications in xx-\left(25-x\right).

$$\frac{x^{2}-25+x}{x}=0$$

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

$$x^{2}-25+x=0$$

Add 25 to both sides. Anything plus zero gives itself.

$$x^{2}+x=25$$

Divide 1, the coefficient of the x term, by 2 to get \frac{1}{2}. Then add the square of \frac{1}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+x+\left(\frac{1}{2}\right)^{2}=25+\left(\frac{1}{2}\right)^{2}$$

Square \frac{1}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+x+\frac{1}{4}=25+\frac{1}{4}$$

Add 25 to \frac{1}{4}.

$$x^{2}+x+\frac{1}{4}=\frac{101}{4}$$

Factor x^{2}+x+\frac{1}{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}.

$$\left(x+\frac{1}{2}\right)^{2}=\frac{101}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{2}\right)^{2}}=\sqrt{\frac{101}{4}}$$

Simplify.

$$x+\frac{1}{2}=\frac{\sqrt{101}}{2} x+\frac{1}{2}=-\frac{\sqrt{101}}{2}$$

Subtract \frac{1}{2} from both sides of the equation.

$$x=\frac{\sqrt{101}-1}{2} x=\frac{-\sqrt{101}-1}{2}$$