Solve x^2+1/x^2-1=13/12

$$\frac{ { x }^{ 2 } +1 }{ { x }^{ 2 } -1 } = \frac{ 13 }{ 12 }$$

Solve for x

$$x=5$$

$$x=-5$$

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 12\left(x-1\right)\left(x+1\right), the least common multiple of x^{2}-1,12.

$$12\left(x^{2}+1\right)=13\left(x-1\right)\left(x+1\right)$$

Use the distributive property to multiply 12 by x^{2}+1.

$$12x^{2}+12=13\left(x-1\right)\left(x+1\right)$$

Use the distributive property to multiply 13 by x-1.

$$12x^{2}+12=\left(13x-13\right)\left(x+1\right)$$

Use the distributive property to multiply 13x-13 by x+1 and combine like terms.

$$12x^{2}+12=13x^{2}-13$$

Subtract 13x^{2} from both sides.

$$12x^{2}+12-13x^{2}=-13$$

Combine 12x^{2} and -13x^{2} to get -x^{2}.

$$-x^{2}+12=-13$$

Subtract 12 from both sides.

$$-x^{2}=-13-12$$

Subtract 12 from -13 to get -25.

$$-x^{2}=-25$$

Divide both sides by -1.

$$x^{2}=\frac{-25}{-1}$$

Fraction \frac{-25}{-1} can be simplified to 25 by removing the negative sign from both the numerator and the denominator.

$$x^{2}=25$$

Take the square root of both sides of the equation.

$$x=5 x=-5$$

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 12\left(x-1\right)\left(x+1\right), the least common multiple of x^{2}-1,12.

$$12\left(x^{2}+1\right)=13\left(x-1\right)\left(x+1\right)$$

Use the distributive property to multiply 12 by x^{2}+1.

$$12x^{2}+12=13\left(x-1\right)\left(x+1\right)$$

Use the distributive property to multiply 13 by x-1.

$$12x^{2}+12=\left(13x-13\right)\left(x+1\right)$$

Use the distributive property to multiply 13x-13 by x+1 and combine like terms.

$$12x^{2}+12=13x^{2}-13$$

Subtract 13x^{2} from both sides.

$$12x^{2}+12-13x^{2}=-13$$

Combine 12x^{2} and -13x^{2} to get -x^{2}.

$$-x^{2}+12=-13$$

Add 13 to both sides.

$$-x^{2}+12+13=0$$

Add 12 and 13 to get 25.

$$-x^{2}+25=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, 0 for b, and 25 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{0±\sqrt{0^{2}-4\left(-1\right)\times 25}}{2\left(-1\right)}$$

Square 0.

$$x=\frac{0±\sqrt{-4\left(-1\right)\times 25}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{0±\sqrt{4\times 25}}{2\left(-1\right)}$$

Multiply 4 times 25.

$$x=\frac{0±\sqrt{100}}{2\left(-1\right)}$$

Take the square root of 100.

$$x=\frac{0±10}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{0±10}{-2}$$

Now solve the equation x=\frac{0±10}{-2} when ± is plus. Divide 10 by -2.

$$x=-5$$

Now solve the equation x=\frac{0±10}{-2} when ± is minus. Divide -10 by -2.

$$x=5$$

The equation is now solved.

$$x=-5 x=5$$