Solve x+1/x=-17

$$x+ \frac{ 1 }{ x } = -17$$

Solve for x

$$x=\frac{\sqrt{285}-17}{2}\approx -0.059028492$$

$$x=\frac{-\sqrt{285}-17}{2}\approx -16.940971508$$

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.

$$xx+1=-17x$$

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

$$x^{2}+1=-17x$$

Add 17x to both sides.

$$x^{2}+1+17x=0$$

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.

$$x^{2}+17x+1=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 17 for b, and 1 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-17±\sqrt{17^{2}-4}}{2}$$

Square 17.

$$x=\frac{-17±\sqrt{289-4}}{2}$$

Add 289 to -4.

$$x=\frac{-17±\sqrt{285}}{2}$$

Now solve the equation x=\frac{-17±\sqrt{285}}{2} when ± is plus. Add -17 to \sqrt{285}.

$$x=\frac{\sqrt{285}-17}{2}$$

Now solve the equation x=\frac{-17±\sqrt{285}}{2} when ± is minus. Subtract \sqrt{285} from -17.

$$x=\frac{-\sqrt{285}-17}{2}$$

The equation is now solved.

$$x=\frac{\sqrt{285}-17}{2} x=\frac{-\sqrt{285}-17}{2}$$

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.

$$xx+1=-17x$$

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

$$x^{2}+1=-17x$$

Add 17x to both sides.

$$x^{2}+1+17x=0$$

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

$$x^{2}+17x=-1$$

Divide 17, the coefficient of the x term, by 2 to get \frac{17}{2}. Then add the square of \frac{17}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+17x+\left(\frac{17}{2}\right)^{2}=-1+\left(\frac{17}{2}\right)^{2}$$

Square \frac{17}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+17x+\frac{289}{4}=-1+\frac{289}{4}$$

Add -1 to \frac{289}{4}.

$$x^{2}+17x+\frac{289}{4}=\frac{285}{4}$$

Factor x^{2}+17x+\frac{289}{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}.

$$\left(x+\frac{17}{2}\right)^{2}=\frac{285}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{17}{2}\right)^{2}}=\sqrt{\frac{285}{4}}$$

Simplify.

$$x+\frac{17}{2}=\frac{\sqrt{285}}{2} x+\frac{17}{2}=-\frac{\sqrt{285}}{2}$$

Subtract \frac{17}{2} from both sides of the equation.

$$x=\frac{\sqrt{285}-17}{2} x=\frac{-\sqrt{285}-17}{2}$$