Solve 1/x-1-x/x+1-2=0

$$\frac { 1 } { x - 1 } - \frac { x } { x + 1 } - 2 = 0$$

Solve for x

$$x = \frac{\sqrt{10} + 1}{3} \approx 1.387425887$$

$$x=\frac{1-\sqrt{10}}{3}\approx -0.72075922$$

Steps Using the Quadratic Formula

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

$$x+1-\left(x-1\right)x+\left(x-1\right)\left(x+1\right)\left(-2\right)=0$$

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

$$x+1-\left(x^{2}-x\right)+\left(x-1\right)\left(x+1\right)\left(-2\right)=0$$

To find the opposite of x^{2}-x, find the opposite of each term.

$$x+1-x^{2}+x+\left(x-1\right)\left(x+1\right)\left(-2\right)=0$$

Combine x and x to get 2x.

$$2x+1-x^{2}+\left(x-1\right)\left(x+1\right)\left(-2\right)=0$$

Use the distributive property to multiply x-1 by x+1 and combine like terms.

$$2x+1-x^{2}+\left(x^{2}-1\right)\left(-2\right)=0$$

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

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

Combine -x^{2} and -2x^{2} to get -3x^{2}.

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

Add 1 and 2 to get 3.

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

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.

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

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

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

Square 2.

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

Multiply -4 times -3.

$$x=\frac{-2±\sqrt{4+12\times 3}}{2\left(-3\right)}$$

Multiply 12 times 3.

$$x=\frac{-2±\sqrt{4+36}}{2\left(-3\right)}$$

Add 4 to 36.

$$x=\frac{-2±\sqrt{40}}{2\left(-3\right)}$$

Take the square root of 40.

$$x=\frac{-2±2\sqrt{10}}{2\left(-3\right)}$$

Multiply 2 times -3.

$$x=\frac{-2±2\sqrt{10}}{-6}$$

Now solve the equation x=\frac{-2±2\sqrt{10}}{-6} when ± is plus. Add -2 to 2\sqrt{10}.

$$x=\frac{2\sqrt{10}-2}{-6}$$

Divide -2+2\sqrt{10} by -6.

$$x=\frac{1-\sqrt{10}}{3}$$

Now solve the equation x=\frac{-2±2\sqrt{10}}{-6} when ± is minus. Subtract 2\sqrt{10} from -2.

$$x=\frac{-2\sqrt{10}-2}{-6}$$

Divide -2-2\sqrt{10} by -6.

$$x=\frac{\sqrt{10}+1}{3}$$

The equation is now solved.

$$x=\frac{1-\sqrt{10}}{3} x=\frac{\sqrt{10}+1}{3}$$

Steps for Completing the Square

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

$$x+1-\left(x-1\right)x+\left(x-1\right)\left(x+1\right)\left(-2\right)=0$$

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

$$x+1-\left(x^{2}-x\right)+\left(x-1\right)\left(x+1\right)\left(-2\right)=0$$

To find the opposite of x^{2}-x, find the opposite of each term.

$$x+1-x^{2}+x+\left(x-1\right)\left(x+1\right)\left(-2\right)=0$$

Combine x and x to get 2x.

$$2x+1-x^{2}+\left(x-1\right)\left(x+1\right)\left(-2\right)=0$$

Use the distributive property to multiply x-1 by x+1 and combine like terms.

$$2x+1-x^{2}+\left(x^{2}-1\right)\left(-2\right)=0$$

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

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

Combine -x^{2} and -2x^{2} to get -3x^{2}.

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

Add 1 and 2 to get 3.

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

Subtract 3 from both sides. Anything subtracted from zero gives its negation.

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

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.

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

Divide both sides by -3.

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

Dividing by -3 undoes the multiplication by -3.

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

Divide 2 by -3.

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

Divide -3 by -3.

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

Divide -\frac{2}{3}, the coefficient of the x term, by 2 to get -\frac{1}{3}. Then add the square of -\frac{1}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square -\frac{1}{3} by squaring both the numerator and the denominator of the fraction.

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

Add 1 to \frac{1}{9}.

$$x^{2}-\frac{2}{3}x+\frac{1}{9}=\frac{10}{9}$$

Factor x^{2}-\frac{2}{3}x+\frac{1}{9}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{1}{3}\right)^{2}=\frac{10}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{3}\right)^{2}}=\sqrt{\frac{10}{9}}$$

Simplify.

$$x-\frac{1}{3}=\frac{\sqrt{10}}{3} x-\frac{1}{3}=-\frac{\sqrt{10}}{3}$$

Add \frac{1}{3} to both sides of the equation.

$$x=\frac{\sqrt{10}+1}{3} x=\frac{1-\sqrt{10}}{3}$$