Solve 3/x^4-1-1/2x^2+2=-1/6-6x^2

$$\frac { 3 } { x ^ { 4 } - 1 } - \frac { 1 } { 2 x ^ { 2 } + 2 } = - \frac { 1 } { 6 - 6 x ^ { 2 } }$$

Solve for x

$$x=\sqrt{5}\approx 2.236067977$$

$$x=-\sqrt{5}\approx -2.236067977$$

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 6\left(x-1\right)\left(x+1\right)\left(x^{2}+1\right), the least common multiple of x^{4}-1,2x^{2}+2,6-6x^{2}.

$$6\times 3-\left(3x^{2}-3\right)=1+x^{2}$$

Multiply 6 and 3 to get 18.

$$18-\left(3x^{2}-3\right)=1+x^{2}$$

To find the opposite of 3x^{2}-3, find the opposite of each term.

$$18-3x^{2}+3=1+x^{2}$$

Add 18 and 3 to get 21.

$$21-3x^{2}=1+x^{2}$$

Subtract x^{2} from both sides.

$$21-3x^{2}-x^{2}=1$$

Combine -3x^{2} and -x^{2} to get -4x^{2}.

$$21-4x^{2}=1$$

Subtract 21 from both sides.

$$-4x^{2}=1-21$$

Subtract 21 from 1 to get -20.

$$-4x^{2}=-20$$

Divide both sides by -4.

$$x^{2}=\frac{-20}{-4}$$

Divide -20 by -4 to get 5.

$$x^{2}=5$$

Take the square root of both sides of the equation.

$$x=\sqrt{5} x=-\sqrt{5}$$

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 6\left(x-1\right)\left(x+1\right)\left(x^{2}+1\right), the least common multiple of x^{4}-1,2x^{2}+2,6-6x^{2}.

$$6\times 3-\left(3x^{2}-3\right)=1+x^{2}$$

Multiply 6 and 3 to get 18.

$$18-\left(3x^{2}-3\right)=1+x^{2}$$

To find the opposite of 3x^{2}-3, find the opposite of each term.

$$18-3x^{2}+3=1+x^{2}$$

Add 18 and 3 to get 21.

$$21-3x^{2}=1+x^{2}$$

Subtract 1 from both sides.

$$21-3x^{2}-1=x^{2}$$

Subtract 1 from 21 to get 20.

$$20-3x^{2}=x^{2}$$

Subtract x^{2} from both sides.

$$20-3x^{2}-x^{2}=0$$

Combine -3x^{2} and -x^{2} to get -4x^{2}.

$$20-4x^{2}=0$$

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.

$$-4x^{2}+20=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -4 for a, 0 for b, and 20 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{0±\sqrt{0^{2}-4\left(-4\right)\times 20}}{2\left(-4\right)}$$

Square 0.

$$x=\frac{0±\sqrt{-4\left(-4\right)\times 20}}{2\left(-4\right)}$$

Multiply -4 times -4.

$$x=\frac{0±\sqrt{16\times 20}}{2\left(-4\right)}$$

Multiply 16 times 20.

$$x=\frac{0±\sqrt{320}}{2\left(-4\right)}$$

Take the square root of 320.

$$x=\frac{0±8\sqrt{5}}{2\left(-4\right)}$$

Multiply 2 times -4.

$$x=\frac{0±8\sqrt{5}}{-8}$$

Now solve the equation x=\frac{0±8\sqrt{5}}{-8} when ± is plus.

$$x=-\sqrt{5}$$

Now solve the equation x=\frac{0±8\sqrt{5}}{-8} when ± is minus.

$$x=\sqrt{5}$$

The equation is now solved.

$$x=-\sqrt{5} x=\sqrt{5}$$