Solve 4/x+1-x+1/x-1=x^2-5/x^2-1

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

Solve for x

$$x=0$$

Steps Using Factoring

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^{2}-1.

$$\left(x-1\right)\times 4-\left(x+1\right)\left(x+1\right)=x^{2}-5$$

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

$$\left(x-1\right)\times 4-\left(x+1\right)^{2}=x^{2}-5$$

Use the distributive property to multiply x-1 by 4.

$$4x-4-\left(x+1\right)^{2}=x^{2}-5$$

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

$$4x-4-\left(x^{2}+2x+1\right)=x^{2}-5$$

To find the opposite of x^{2}+2x+1, find the opposite of each term.

$$4x-4-x^{2}-2x-1=x^{2}-5$$

Combine 4x and -2x to get 2x.

$$2x-4-x^{2}-1=x^{2}-5$$

Subtract 1 from -4 to get -5.

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

Subtract x^{2} from both sides.

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

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

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

Add 5 to both sides.

$$2x-5-2x^{2}+5=0$$

Add -5 and 5 to get 0.

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

Factor out x.

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

To find equation solutions, solve x=0 and 2-2x=0.

$$x=0 x=1$$

Variable x cannot be equal to 1.

$$x=0$$

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,x^{2}-1.

$$\left(x-1\right)\times 4-\left(x+1\right)\left(x+1\right)=x^{2}-5$$

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

$$\left(x-1\right)\times 4-\left(x+1\right)^{2}=x^{2}-5$$

Use the distributive property to multiply x-1 by 4.

$$4x-4-\left(x+1\right)^{2}=x^{2}-5$$

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

$$4x-4-\left(x^{2}+2x+1\right)=x^{2}-5$$

To find the opposite of x^{2}+2x+1, find the opposite of each term.

$$4x-4-x^{2}-2x-1=x^{2}-5$$

Combine 4x and -2x to get 2x.

$$2x-4-x^{2}-1=x^{2}-5$$

Subtract 1 from -4 to get -5.

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

Subtract x^{2} from both sides.

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

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

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

Add 5 to both sides.

$$2x-5-2x^{2}+5=0$$

Add -5 and 5 to get 0.

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

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.

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

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

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

Take the square root of 2^{2}.

$$x=\frac{-2±2}{2\left(-2\right)}$$

Multiply 2 times -2.

$$x=\frac{-2±2}{-4}$$

Now solve the equation x=\frac{-2±2}{-4} when ± is plus. Add -2 to 2.

$$x=\frac{0}{-4}$$

Divide 0 by -4.

$$x=0$$

Now solve the equation x=\frac{-2±2}{-4} when ± is minus. Subtract 2 from -2.

$$x=-\frac{4}{-4}$$

Divide -4 by -4.

$$x=1$$

The equation is now solved.

$$x=0 x=1$$

Variable x cannot be equal to 1.

$$x=0$$

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^{2}-1.

$$\left(x-1\right)\times 4-\left(x+1\right)\left(x+1\right)=x^{2}-5$$

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

$$\left(x-1\right)\times 4-\left(x+1\right)^{2}=x^{2}-5$$

Use the distributive property to multiply x-1 by 4.

$$4x-4-\left(x+1\right)^{2}=x^{2}-5$$

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

$$4x-4-\left(x^{2}+2x+1\right)=x^{2}-5$$

To find the opposite of x^{2}+2x+1, find the opposite of each term.

$$4x-4-x^{2}-2x-1=x^{2}-5$$

Combine 4x and -2x to get 2x.

$$2x-4-x^{2}-1=x^{2}-5$$

Subtract 1 from -4 to get -5.

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

Subtract x^{2} from both sides.

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

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

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

Add 5 to both sides.

$$2x-2x^{2}=-5+5$$

Add -5 and 5 to get 0.

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

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.

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

Divide both sides by -2.

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

Dividing by -2 undoes the multiplication by -2.

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

Divide 2 by -2.

$$x^{2}-x=\frac{0}{-2}$$

Divide 0 by -2.

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

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

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

Square -\frac{1}{2} by squaring both the numerator and the denominator of the fraction.

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

Factor x^{2}-x+\frac{1}{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}.

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

Take the square root of both sides of the equation.

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

Simplify.

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

Add \frac{1}{2} to both sides of the equation.

$$x=1 x=0$$

Variable x cannot be equal to 1.

$$x=0$$