Solve x+1/x^2-x=2

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

Solve for x

$$x = \frac{\sqrt{17} + 3}{4} \approx 1.780776406$$

$$x=\frac{3-\sqrt{17}}{4}\approx -0.280776406$$

Steps Using the Quadratic Formula

Variable x cannot be equal to any of the values 0,1 since division by zero is not defined. Multiply both sides of the equation by x\left(x-1\right).

$$x+1=2x\left(x-1\right)$$

Use the distributive property to multiply 2x by x-1.

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

Subtract 2x^{2} from both sides.

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

Add 2x to both sides.

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

Combine x and 2x to get 3x.

$$3x+1-2x^{2}=0$$

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.

$$-2x^{2}+3x+1=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -2 for a, 3 for b, and 1 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-3±\sqrt{3^{2}-4\left(-2\right)}}{2\left(-2\right)}$$

Square 3.

$$x=\frac{-3±\sqrt{9-4\left(-2\right)}}{2\left(-2\right)}$$

Multiply -4 times -2.

$$x=\frac{-3±\sqrt{9+8}}{2\left(-2\right)}$$

Add 9 to 8.

$$x=\frac{-3±\sqrt{17}}{2\left(-2\right)}$$

Multiply 2 times -2.

$$x=\frac{-3±\sqrt{17}}{-4}$$

Now solve the equation x=\frac{-3±\sqrt{17}}{-4} when ± is plus. Add -3 to \sqrt{17}.

$$x=\frac{\sqrt{17}-3}{-4}$$

Divide -3+\sqrt{17} by -4.

$$x=\frac{3-\sqrt{17}}{4}$$

Now solve the equation x=\frac{-3±\sqrt{17}}{-4} when ± is minus. Subtract \sqrt{17} from -3.

$$x=\frac{-\sqrt{17}-3}{-4}$$

Divide -3-\sqrt{17} by -4.

$$x=\frac{\sqrt{17}+3}{4}$$

The equation is now solved.

$$x=\frac{3-\sqrt{17}}{4} x=\frac{\sqrt{17}+3}{4}$$

Steps for Completing the Square

Variable x cannot be equal to any of the values 0,1 since division by zero is not defined. Multiply both sides of the equation by x\left(x-1\right).

$$x+1=2x\left(x-1\right)$$

Use the distributive property to multiply 2x by x-1.

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

Subtract 2x^{2} from both sides.

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

Add 2x to both sides.

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

Combine x and 2x to get 3x.

$$3x+1-2x^{2}=0$$

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

$$3x-2x^{2}=-1$$

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.

$$-2x^{2}+3x=-1$$

Divide both sides by -2.

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

Dividing by -2 undoes the multiplication by -2.

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

Divide 3 by -2.

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

Divide -1 by -2.

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

Divide -\frac{3}{2}, the coefficient of the x term, by 2 to get -\frac{3}{4}. Then add the square of -\frac{3}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{3}{2}x+\left(-\frac{3}{4}\right)^{2}=\frac{1}{2}+\left(-\frac{3}{4}\right)^{2}$$

Square -\frac{3}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{3}{2}x+\frac{9}{16}=\frac{1}{2}+\frac{9}{16}$$

Add \frac{1}{2} to \frac{9}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{3}{2}x+\frac{9}{16}=\frac{17}{16}$$

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

$$\left(x-\frac{3}{4}\right)^{2}=\frac{17}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{3}{4}\right)^{2}}=\sqrt{\frac{17}{16}}$$

Simplify.

$$x-\frac{3}{4}=\frac{\sqrt{17}}{4} x-\frac{3}{4}=-\frac{\sqrt{17}}{4}$$

Add \frac{3}{4} to both sides of the equation.

$$x=\frac{\sqrt{17}+3}{4} x=\frac{3-\sqrt{17}}{4}$$