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

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

Solve for x

$$x=\frac{\sqrt{5}-1}{2}\approx 0.618033989$$

$$x=\frac{-\sqrt{5}-1}{2}\approx -1.618033989$$

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^{2}-1,x-1.

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

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

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

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

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

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

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

Subtract 1 from 2 to get 1.

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

Consider \left(x-1\right)\left(x+1\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 1.

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

Subtract x^{2} from both sides.

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

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

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

Add 1 to both sides.

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

Add 1 and 1 to get 2.

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

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.

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

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

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

Square -2.

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

Multiply -4 times -2.

$$x=\frac{-\left(-2\right)±\sqrt{4+8\times 2}}{2\left(-2\right)}$$

Multiply 8 times 2.

$$x=\frac{-\left(-2\right)±\sqrt{4+16}}{2\left(-2\right)}$$

Add 4 to 16.

$$x=\frac{-\left(-2\right)±\sqrt{20}}{2\left(-2\right)}$$

Take the square root of 20.

$$x=\frac{-\left(-2\right)±2\sqrt{5}}{2\left(-2\right)}$$

The opposite of -2 is 2.

$$x=\frac{2±2\sqrt{5}}{2\left(-2\right)}$$

Multiply 2 times -2.

$$x=\frac{2±2\sqrt{5}}{-4}$$

Now solve the equation x=\frac{2±2\sqrt{5}}{-4} when ± is plus. Add 2 to 2\sqrt{5}.

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

Divide 2+2\sqrt{5} by -4.

$$x=\frac{-\sqrt{5}-1}{2}$$

Now solve the equation x=\frac{2±2\sqrt{5}}{-4} when ± is minus. Subtract 2\sqrt{5} from 2.

$$x=\frac{2-2\sqrt{5}}{-4}$$

Divide 2-2\sqrt{5} by -4.

$$x=\frac{\sqrt{5}-1}{2}$$

The equation is now solved.

$$x=\frac{-\sqrt{5}-1}{2} x=\frac{\sqrt{5}-1}{2}$$

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^{2}-1,x-1.

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

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

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

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

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

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

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

Subtract 1 from 2 to get 1.

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

Consider \left(x-1\right)\left(x+1\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 1.

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

Subtract x^{2} from both sides.

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

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

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

Subtract 1 from both sides.

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

Subtract 1 from -1 to get -2.

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

Divide both sides by -2.

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

Dividing by -2 undoes the multiplication by -2.

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

Divide -2 by -2.

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

Divide -2 by -2.

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

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.

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

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

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

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

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

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}.

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

Take the square root of both sides of the equation.

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

Simplify.

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

Subtract \frac{1}{2} from both sides of the equation.

$$x=\frac{\sqrt{5}-1}{2} x=\frac{-\sqrt{5}-1}{2}$$