Solve x^{2}+1/x^2=x^2+1/x^2-2+2=(x-1/x)^2+2

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

Evaluate (complex solution)

Short Solution Steps

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

$$\frac{x^{2}x^{2}}{x^{2}}+\frac{1}{x^{2}}=x^{2}+\frac{1}{x^{2}}-2+2\text{ and }x^{2}+\frac{1}{x^{2}}-2+2=\left(x-\frac{1}{x}\right)^{2}+2$$

Since \frac{x^{2}x^{2}}{x^{2}} and \frac{1}{x^{2}} have the same denominator, add them by adding their numerators.

$$\frac{x^{2}x^{2}+1}{x^{2}}=x^{2}+\frac{1}{x^{2}}-2+2\text{ and }x^{2}+\frac{1}{x^{2}}-2+2=\left(x-\frac{1}{x}\right)^{2}+2$$

Do the multiplications in x^{2}x^{2}+1.

$$\frac{x^{4}+1}{x^{2}}=x^{2}+\frac{1}{x^{2}}-2+2\text{ and }x^{2}+\frac{1}{x^{2}}-2+2=\left(x-\frac{1}{x}\right)^{2}+2$$

Add -2 and 2 to get 0.

$$\frac{x^{4}+1}{x^{2}}=x^{2}+\frac{1}{x^{2}}\text{ and }x^{2}+\frac{1}{x^{2}}-2+2=\left(x-\frac{1}{x}\right)^{2}+2$$

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

$$\frac{x^{4}+1}{x^{2}}=\frac{x^{2}x^{2}}{x^{2}}+\frac{1}{x^{2}}\text{ and }x^{2}+\frac{1}{x^{2}}-2+2=\left(x-\frac{1}{x}\right)^{2}+2$$

Since \frac{x^{2}x^{2}}{x^{2}} and \frac{1}{x^{2}} have the same denominator, add them by adding their numerators.

$$\frac{x^{4}+1}{x^{2}}=\frac{x^{2}x^{2}+1}{x^{2}}\text{ and }x^{2}+\frac{1}{x^{2}}-2+2=\left(x-\frac{1}{x}\right)^{2}+2$$

Do the multiplications in x^{2}x^{2}+1.

$$\frac{x^{4}+1}{x^{2}}=\frac{x^{4}+1}{x^{2}}\text{ and }x^{2}+\frac{1}{x^{2}}-2+2=\left(x-\frac{1}{x}\right)^{2}+2$$

Add -2 and 2 to get 0.

$$\frac{x^{4}+1}{x^{2}}=\frac{x^{4}+1}{x^{2}}\text{ and }x^{2}+\frac{1}{x^{2}}=\left(x-\frac{1}{x}\right)^{2}+2$$

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

$$\frac{x^{4}+1}{x^{2}}=\frac{x^{4}+1}{x^{2}}\text{ and }\frac{x^{2}x^{2}}{x^{2}}+\frac{1}{x^{2}}=\left(x-\frac{1}{x}\right)^{2}+2$$

Since \frac{x^{2}x^{2}}{x^{2}} and \frac{1}{x^{2}} have the same denominator, add them by adding their numerators.

$$\frac{x^{4}+1}{x^{2}}=\frac{x^{4}+1}{x^{2}}\text{ and }\frac{x^{2}x^{2}+1}{x^{2}}=\left(x-\frac{1}{x}\right)^{2}+2$$

Do the multiplications in x^{2}x^{2}+1.

$$\frac{x^{4}+1}{x^{2}}=\frac{x^{4}+1}{x^{2}}\text{ and }\frac{x^{4}+1}{x^{2}}=\left(x-\frac{1}{x}\right)^{2}+2$$

To add or subtract expressions, expand them to make their denominators the same. Multiply x times \frac{x}{x}.

$$\frac{x^{4}+1}{x^{2}}=\frac{x^{4}+1}{x^{2}}\text{ and }\frac{x^{4}+1}{x^{2}}=\left(\frac{xx}{x}-\frac{1}{x}\right)^{2}+2$$

Since \frac{xx}{x} and \frac{1}{x} have the same denominator, subtract them by subtracting their numerators.

$$\frac{x^{4}+1}{x^{2}}=\frac{x^{4}+1}{x^{2}}\text{ and }\frac{x^{4}+1}{x^{2}}=\left(\frac{xx-1}{x}\right)^{2}+2$$

Do the multiplications in xx-1.

$$\frac{x^{4}+1}{x^{2}}=\frac{x^{4}+1}{x^{2}}\text{ and }\frac{x^{4}+1}{x^{2}}=\left(\frac{x^{2}-1}{x}\right)^{2}+2$$

To raise \frac{x^{2}-1}{x} to a power, raise both numerator and denominator to the power and then divide.

$$\frac{x^{4}+1}{x^{2}}=\frac{x^{4}+1}{x^{2}}\text{ and }\frac{x^{4}+1}{x^{2}}=\frac{\left(x^{2}-1\right)^{2}}{x^{2}}+2$$

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

$$\frac{x^{4}+1}{x^{2}}=\frac{x^{4}+1}{x^{2}}\text{ and }\frac{x^{4}+1}{x^{2}}=\frac{\left(x^{2}-1\right)^{2}}{x^{2}}+\frac{2x^{2}}{x^{2}}$$

Since \frac{\left(x^{2}-1\right)^{2}}{x^{2}} and \frac{2x^{2}}{x^{2}} have the same denominator, add them by adding their numerators.

$$\frac{x^{4}+1}{x^{2}}=\frac{x^{4}+1}{x^{2}}\text{ and }\frac{x^{4}+1}{x^{2}}=\frac{\left(x^{2}-1\right)^{2}+2x^{2}}{x^{2}}$$

Do the multiplications in \left(x^{2}-1\right)^{2}+2x^{2}.

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

Combine like terms in x^{4}-2x^{2}+1+2x^{2}.

$$\frac{x^{4}+1}{x^{2}}=\frac{x^{4}+1}{x^{2}}\text{ and }\frac{x^{4}+1}{x^{2}}=\frac{x^{4}+1}{x^{2}}$$

Subtract \frac{x^{4}+1}{x^{2}} from both sides.

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

Subtract \frac{x^{4}+1}{x^{2}} from \frac{x^{4}+1}{x^{2}} to get 0.

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

Compare 0 and 0.

$$\text{true}\text{ and }\frac{x^{4}+1}{x^{2}}=\frac{x^{4}+1}{x^{2}}$$

Subtract \frac{x^{4}+1}{x^{2}} from both sides.

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

Subtract \frac{x^{4}+1}{x^{2}} from \frac{x^{4}+1}{x^{2}} to get 0.

$$\text{true}\text{ and }0=0$$

Compare 0 and 0.

$$\text{true}\text{ and }\text{true}$$

The conjunction of \text{true} and \text{true} is \text{true}.

$$\text{true}$$

Solve for x

$$x\neq 0$$