Solve 4/x^2-1+2/1-2=0

$$\frac { 4 } { x ^ { 2 } - 1 } + \frac { 2 } { 1 - 2 } = 0$$

Solve for x

$$x=\sqrt{3}\approx 1.732050808$$

$$x=-\sqrt{3}\approx -1.732050808$$

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).

$$4+\left(-x^{2}+1\right)\times 2=0$$

Use the distributive property to multiply -x^{2}+1 by 2.

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

Add 4 and 2 to get 6.

$$6-2x^{2}=0$$

Subtract 6 from both sides. Anything subtracted from zero gives its negation.

$$-2x^{2}=-6$$

Divide both sides by -2.

$$x^{2}=\frac{-6}{-2}$$

Divide -6 by -2 to get 3.

$$x^{2}=3$$

Take the square root of both sides of the equation.

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

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).

$$4+\left(-x^{2}+1\right)\times 2=0$$

Use the distributive property to multiply -x^{2}+1 by 2.

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

Add 4 and 2 to get 6.

$$6-2x^{2}=0$$

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.

$$-2x^{2}+6=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -2 for a, 0 for b, and 6 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{0±\sqrt{0^{2}-4\left(-2\right)\times 6}}{2\left(-2\right)}$$

Square 0.

$$x=\frac{0±\sqrt{-4\left(-2\right)\times 6}}{2\left(-2\right)}$$

Multiply -4 times -2.

$$x=\frac{0±\sqrt{8\times 6}}{2\left(-2\right)}$$

Multiply 8 times 6.

$$x=\frac{0±\sqrt{48}}{2\left(-2\right)}$$

Take the square root of 48.

$$x=\frac{0±4\sqrt{3}}{2\left(-2\right)}$$

Multiply 2 times -2.

$$x=\frac{0±4\sqrt{3}}{-4}$$

Now solve the equation x=\frac{0±4\sqrt{3}}{-4} when ± is plus.

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

Now solve the equation x=\frac{0±4\sqrt{3}}{-4} when ± is minus.

$$x=\sqrt{3}$$

The equation is now solved.

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