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

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

Solve for x

$$x=-1$$

Steps by Finding Square Root

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

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

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

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

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

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

Combine x and -x to get 0.

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

Subtract 2 from both sides. Anything subtracted from zero gives its negation.

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

Divide both sides by -2.

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

Divide -2 by -2 to get 1.

$$x^{2}=1$$

Take the square root of both sides of the equation.

$$x=1 x=-1$$

Variable x cannot be equal to 1.

$$x=-1$$

Steps Using the Quadratic Formula

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

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

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

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

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

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

Combine x and -x to get 0.

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

Quadratic equations like this one, with an x^{2} term but no x term, can still be solved using the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}, once they are put in standard form: ax^{2}+bx+c=0.

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

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

$$x=\frac{0±\sqrt{0^{2}-4\left(-2\right)\times 2}}{2\left(-2\right)}$$

Square 0.

$$x=\frac{0±\sqrt{-4\left(-2\right)\times 2}}{2\left(-2\right)}$$

Multiply -4 times -2.

$$x=\frac{0±\sqrt{8\times 2}}{2\left(-2\right)}$$

Multiply 8 times 2.

$$x=\frac{0±\sqrt{16}}{2\left(-2\right)}$$

Take the square root of 16.

$$x=\frac{0±4}{2\left(-2\right)}$$

Multiply 2 times -2.

$$x=\frac{0±4}{-4}$$

Now solve the equation x=\frac{0±4}{-4} when ± is plus. Divide 4 by -4.

$$x=-1$$

Now solve the equation x=\frac{0±4}{-4} when ± is minus. Divide -4 by -4.

$$x=1$$

The equation is now solved.

$$x=-1 x=1$$

Variable x cannot be equal to 1.

$$x=-1$$