Solve (x/x-2)^2+(x/x+2)^2=2

$$( \frac { x } { x - 2 } ) ^ { 2 } + ( \frac { x } { x + 2 } ) ^ { 2 } = 2$$

Solve for x

$$x = \frac{2 \sqrt{3}}{3} \approx 1.154700538$$

$$x = -\frac{2 \sqrt{3}}{3} \approx -1.154700538$$

Steps by Finding Square Root

To raise \frac{x}{x-2} to a power, raise both numerator and denominator to the power and then divide.

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

To raise \frac{x}{x+2} to a power, raise both numerator and denominator to the power and then divide.

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

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of \left(x-2\right)^{2} and \left(x+2\right)^{2} is \left(x-2\right)^{2}\left(x+2\right)^{2}. Multiply \frac{x^{2}}{\left(x-2\right)^{2}} times \frac{\left(x+2\right)^{2}}{\left(x+2\right)^{2}}. Multiply \frac{x^{2}}{\left(x+2\right)^{2}} times \frac{\left(x-2\right)^{2}}{\left(x-2\right)^{2}}.

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

Since \frac{x^{2}\left(x+2\right)^{2}}{\left(x-2\right)^{2}\left(x+2\right)^{2}} and \frac{x^{2}\left(x-2\right)^{2}}{\left(x-2\right)^{2}\left(x+2\right)^{2}} have the same denominator, add them by adding their numerators.

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

Do the multiplications in x^{2}\left(x+2\right)^{2}+x^{2}\left(x-2\right)^{2}.

$$\frac{x^{4}+4x^{3}+4x^{2}+x^{4}-4x^{3}+4x^{2}}{\left(x-2\right)^{2}\left(x+2\right)^{2}}=2$$

Combine like terms in x^{4}+4x^{3}+4x^{2}+x^{4}-4x^{3}+4x^{2}.

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

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

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

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

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

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

$$2x^{4}+8x^{2}=2\left(x-2\right)^{2}\left(x+2\right)^{2}$$

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

$$2x^{4}+8x^{2}=2\left(x^{2}-4x+4\right)\left(x+2\right)^{2}$$

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

$$2x^{4}+8x^{2}=2\left(x^{2}-4x+4\right)\left(x^{2}+4x+4\right)$$

Use the distributive property to multiply 2 by x^{2}-4x+4.

$$2x^{4}+8x^{2}=\left(2x^{2}-8x+8\right)\left(x^{2}+4x+4\right)$$

Use the distributive property to multiply 2x^{2}-8x+8 by x^{2}+4x+4 and combine like terms.

$$2x^{4}+8x^{2}=2x^{4}-16x^{2}+32$$

Subtract 2x^{4} from both sides.

$$2x^{4}+8x^{2}-2x^{4}=-16x^{2}+32$$

Combine 2x^{4} and -2x^{4} to get 0.

$$8x^{2}=-16x^{2}+32$$

Add 16x^{2} to both sides.

$$8x^{2}+16x^{2}=32$$

Combine 8x^{2} and 16x^{2} to get 24x^{2}.

$$24x^{2}=32$$

Divide both sides by 24.

$$x^{2}=\frac{32}{24}$$

Reduce the fraction \frac{32}{24} to lowest terms by extracting and canceling out 8.

$$x^{2}=\frac{4}{3}$$

Take the square root of both sides of the equation.

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

Steps Using the Quadratic Formula

To raise \frac{x}{x-2} to a power, raise both numerator and denominator to the power and then divide.

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

To raise \frac{x}{x+2} to a power, raise both numerator and denominator to the power and then divide.

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

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of \left(x-2\right)^{2} and \left(x+2\right)^{2} is \left(x-2\right)^{2}\left(x+2\right)^{2}. Multiply \frac{x^{2}}{\left(x-2\right)^{2}} times \frac{\left(x+2\right)^{2}}{\left(x+2\right)^{2}}. Multiply \frac{x^{2}}{\left(x+2\right)^{2}} times \frac{\left(x-2\right)^{2}}{\left(x-2\right)^{2}}.

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

Since \frac{x^{2}\left(x+2\right)^{2}}{\left(x-2\right)^{2}\left(x+2\right)^{2}} and \frac{x^{2}\left(x-2\right)^{2}}{\left(x-2\right)^{2}\left(x+2\right)^{2}} have the same denominator, add them by adding their numerators.

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

Do the multiplications in x^{2}\left(x+2\right)^{2}+x^{2}\left(x-2\right)^{2}.

$$\frac{x^{4}+4x^{3}+4x^{2}+x^{4}-4x^{3}+4x^{2}}{\left(x-2\right)^{2}\left(x+2\right)^{2}}=2$$

Combine like terms in x^{4}+4x^{3}+4x^{2}+x^{4}-4x^{3}+4x^{2}.

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

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

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

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

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

Subtract 2 from both sides.

$$\frac{2x^{4}+8x^{2}}{\left(x^{2}-4x+4\right)\left(x^{2}+4x+4\right)}-2=0$$

Use the distributive property to multiply x^{2}-4x+4 by x^{2}+4x+4 and combine like terms.

$$\frac{2x^{4}+8x^{2}}{x^{4}-8x^{2}+16}-2=0$$

Factor x^{4}-8x^{2}+16.

$$\frac{2x^{4}+8x^{2}}{\left(x-2\right)^{2}\left(x+2\right)^{2}}-2=0$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 2 times \frac{\left(x-2\right)^{2}\left(x+2\right)^{2}}{\left(x-2\right)^{2}\left(x+2\right)^{2}}.

$$\frac{2x^{4}+8x^{2}}{\left(x-2\right)^{2}\left(x+2\right)^{2}}-\frac{2\left(x-2\right)^{2}\left(x+2\right)^{2}}{\left(x-2\right)^{2}\left(x+2\right)^{2}}=0$$

Since \frac{2x^{4}+8x^{2}}{\left(x-2\right)^{2}\left(x+2\right)^{2}} and \frac{2\left(x-2\right)^{2}\left(x+2\right)^{2}}{\left(x-2\right)^{2}\left(x+2\right)^{2}} have the same denominator, subtract them by subtracting their numerators.

$$\frac{2x^{4}+8x^{2}-2\left(x-2\right)^{2}\left(x+2\right)^{2}}{\left(x-2\right)^{2}\left(x+2\right)^{2}}=0$$

Do the multiplications in 2x^{4}+8x^{2}-2\left(x-2\right)^{2}\left(x+2\right)^{2}.

$$\frac{2x^{4}+8x^{2}-2x^{4}-8x^{3}-8x^{2}+8x^{3}+32x^{2}+32x-8x^{2}-32x-32}{\left(x-2\right)^{2}\left(x+2\right)^{2}}=0$$

Combine like terms in 2x^{4}+8x^{2}-2x^{4}-8x^{3}-8x^{2}+8x^{3}+32x^{2}+32x-8x^{2}-32x-32.

$$\frac{24x^{2}-32}{\left(x-2\right)^{2}\left(x+2\right)^{2}}=0$$

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

$$24x^{2}-32=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 24 for a, 0 for b, and -32 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{0±\sqrt{0^{2}-4\times 24\left(-32\right)}}{2\times 24}$$

Square 0.

$$x=\frac{0±\sqrt{-4\times 24\left(-32\right)}}{2\times 24}$$

Multiply -4 times 24.

$$x=\frac{0±\sqrt{-96\left(-32\right)}}{2\times 24}$$

Multiply -96 times -32.

$$x=\frac{0±\sqrt{3072}}{2\times 24}$$

Take the square root of 3072.

$$x=\frac{0±32\sqrt{3}}{2\times 24}$$

Multiply 2 times 24.

$$x=\frac{0±32\sqrt{3}}{48}$$

Now solve the equation x=\frac{0±32\sqrt{3}}{48} when ± is plus.

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

Now solve the equation x=\frac{0±32\sqrt{3}}{48} when ± is minus.

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

The equation is now solved.

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