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

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

Solve for x

$$x=0$$

Steps by Finding Square Root

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.

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

Multiply x-1 and x-1 to get \left(x-1\right)^{2}.

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

Multiply x+1 and x+1 to get \left(x+1\right)^{2}.

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

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-1\right)^{2}.

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

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+1\right)^{2}.

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

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

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

Combine -2x and 2x to get 0.

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

Add 1 and 1 to get 2.

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

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

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

Use the distributive property to multiply -2x+2 by x+1 and combine like terms.

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

Add 2x^{2} to both sides.

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

Combine 2x^{2} and 2x^{2} to get 4x^{2}.

$$4x^{2}+2=2$$

Subtract 2 from both sides.

$$4x^{2}=2-2$$

Subtract 2 from 2 to get 0.

$$4x^{2}=0$$

Divide both sides by 4. Zero divided by any non-zero number gives zero.

$$x^{2}=0$$

Take the square root of both sides of the equation.

$$x=0 x=0$$

The equation is now solved. Solutions are the same.

$$x=0$$

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.

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

Multiply x-1 and x-1 to get \left(x-1\right)^{2}.

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

Multiply x+1 and x+1 to get \left(x+1\right)^{2}.

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

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-1\right)^{2}.

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

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+1\right)^{2}.

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

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

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

Combine -2x and 2x to get 0.

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

Add 1 and 1 to get 2.

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

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

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

Use the distributive property to multiply -2x+2 by x+1 and combine like terms.

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

Add 2x^{2} to both sides.

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

Combine 2x^{2} and 2x^{2} to get 4x^{2}.

$$4x^{2}+2=2$$

Subtract 2 from both sides.

$$4x^{2}+2-2=0$$

Subtract 2 from 2 to get 0.

$$4x^{2}=0$$

Divide both sides by 4. Zero divided by any non-zero number gives zero.

$$x^{2}=0$$

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

$$x=\frac{0±\sqrt{0^{2}}}{2}$$

Take the square root of 0^{2}.

$$x=\frac{0±0}{2}$$

Divide 0 by 2.

$$x=0$$