Solve 5x-1/x=15

$$5x- \frac{ 1 }{ x } =15$$

Solve for x

$$x=\frac{7\sqrt{5}}{10}+\frac{3}{2}\approx 3.065247584$$

$$x=-\frac{7\sqrt{5}}{10}+\frac{3}{2}\approx -0.065247584$$

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.

$$5xx-1=15x$$

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

$$5x^{2}-1=15x$$

Subtract 15x from both sides.

$$5x^{2}-1-15x=0$$

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.

$$5x^{2}-15x-1=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 5 for a, -15 for b, and -1 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-15\right)±\sqrt{\left(-15\right)^{2}-4\times 5\left(-1\right)}}{2\times 5}$$

Square -15.

$$x=\frac{-\left(-15\right)±\sqrt{225-4\times 5\left(-1\right)}}{2\times 5}$$

Multiply -4 times 5.

$$x=\frac{-\left(-15\right)±\sqrt{225-20\left(-1\right)}}{2\times 5}$$

Multiply -20 times -1.

$$x=\frac{-\left(-15\right)±\sqrt{225+20}}{2\times 5}$$

Add 225 to 20.

$$x=\frac{-\left(-15\right)±\sqrt{245}}{2\times 5}$$

Take the square root of 245.

$$x=\frac{-\left(-15\right)±7\sqrt{5}}{2\times 5}$$

The opposite of -15 is 15.

$$x=\frac{15±7\sqrt{5}}{2\times 5}$$

Multiply 2 times 5.

$$x=\frac{15±7\sqrt{5}}{10}$$

Now solve the equation x=\frac{15±7\sqrt{5}}{10} when ± is plus. Add 15 to 7\sqrt{5}.

$$x=\frac{7\sqrt{5}+15}{10}$$

Divide 15+7\sqrt{5} by 10.

$$x=\frac{7\sqrt{5}}{10}+\frac{3}{2}$$

Now solve the equation x=\frac{15±7\sqrt{5}}{10} when ± is minus. Subtract 7\sqrt{5} from 15.

$$x=\frac{15-7\sqrt{5}}{10}$$

Divide 15-7\sqrt{5} by 10.

$$x=-\frac{7\sqrt{5}}{10}+\frac{3}{2}$$

The equation is now solved.

$$x=\frac{7\sqrt{5}}{10}+\frac{3}{2} x=-\frac{7\sqrt{5}}{10}+\frac{3}{2}$$

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.

$$5xx-1=15x$$

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

$$5x^{2}-1=15x$$

Subtract 15x from both sides.

$$5x^{2}-1-15x=0$$

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

$$5x^{2}-15x=1$$

Divide both sides by 5.

$$\frac{5x^{2}-15x}{5}=\frac{1}{5}$$

Dividing by 5 undoes the multiplication by 5.

$$x^{2}+\left(-\frac{15}{5}\right)x=\frac{1}{5}$$

Divide -15 by 5.

$$x^{2}-3x=\frac{1}{5}$$

Divide -3, the coefficient of the x term, by 2 to get -\frac{3}{2}. Then add the square of -\frac{3}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-3x+\left(-\frac{3}{2}\right)^{2}=\frac{1}{5}+\left(-\frac{3}{2}\right)^{2}$$

Square -\frac{3}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-3x+\frac{9}{4}=\frac{1}{5}+\frac{9}{4}$$

Add \frac{1}{5} to \frac{9}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-3x+\frac{9}{4}=\frac{49}{20}$$

Factor x^{2}-3x+\frac{9}{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}.

$$\left(x-\frac{3}{2}\right)^{2}=\frac{49}{20}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{3}{2}\right)^{2}}=\sqrt{\frac{49}{20}}$$

Simplify.

$$x-\frac{3}{2}=\frac{7\sqrt{5}}{10} x-\frac{3}{2}=-\frac{7\sqrt{5}}{10}$$

Add \frac{3}{2} to both sides of the equation.

$$x=\frac{7\sqrt{5}}{10}+\frac{3}{2} x=-\frac{7\sqrt{5}}{10}+\frac{3}{2}$$