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

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

Solve for x

$$x=\sqrt{3}+1\approx 2.732050808$$

$$x=1-\sqrt{3}\approx -0.732050808$$

Steps Using the Quadratic Formula

Variable x cannot be equal to any of the values -1,0,1 since division by zero is not defined. Multiply both sides of the equation by x\left(x-1\right)\left(x+1\right), the least common multiple of x^{2}-x,1-x^{2},x.

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

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

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

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

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

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

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

Combine x^{2} and -x^{2} to get 0.

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

Subtract x^{2} from both sides.

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

Add 1 to both sides.

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

Add 1 and 1 to get 2.

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

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.

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

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

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

Square 2.

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

Multiply -4 times -1.

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

Multiply 4 times 2.

$$x=\frac{-2±\sqrt{4+8}}{2\left(-1\right)}$$

Add 4 to 8.

$$x=\frac{-2±\sqrt{12}}{2\left(-1\right)}$$

Take the square root of 12.

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

Multiply 2 times -1.

$$x=\frac{-2±2\sqrt{3}}{-2}$$

Now solve the equation x=\frac{-2±2\sqrt{3}}{-2} when ± is plus. Add -2 to 2\sqrt{3}.

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

Divide -2+2\sqrt{3} by -2.

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

Now solve the equation x=\frac{-2±2\sqrt{3}}{-2} when ± is minus. Subtract 2\sqrt{3} from -2.

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

Divide -2-2\sqrt{3} by -2.

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

The equation is now solved.

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

Steps for Completing the Square

Variable x cannot be equal to any of the values -1,0,1 since division by zero is not defined. Multiply both sides of the equation by x\left(x-1\right)\left(x+1\right), the least common multiple of x^{2}-x,1-x^{2},x.

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

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

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

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

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

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

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

Combine x^{2} and -x^{2} to get 0.

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

Subtract x^{2} from both sides.

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

Subtract 1 from both sides.

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

Subtract 1 from -1 to get -2.

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

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.

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

Divide both sides by -1.

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

Dividing by -1 undoes the multiplication by -1.

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

Divide 2 by -1.

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

Divide -2 by -1.

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

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

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

Add 2 to 1.

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

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

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

Take the square root of both sides of the equation.

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

Simplify.

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

Add 1 to both sides of the equation.

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