Evaluate square root of minus one to the power of two plus ( square root of two) to the power of two

$\sqrt{- 1^{2} + {(\sqrt{2})}^{2}}$

Clarify the expression.

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One raised to any power equals one.

$\sqrt{- 1 \cdot 1 + {(\sqrt{2})}^{2}}$

First, multiply $- 1$ by $1$ .

$\sqrt{- 1 + {(\sqrt{2})}^{2}}$

Should be rewritten ${\sqrt{2}}^{2}$ as $2$ .

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Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{2}$ as $2^{\frac{1}{2}}$ .

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

Apply the power rule when multiplying the exponents, ${(a^{m})}^{n} = a^{m n}$ .

$\sqrt{- 1 + 2^{\frac{1}{2} \cdot 2}}$

Add $\frac{1}{2}$ and $2$ .

$\sqrt{- 1 + 2^{\frac{2}{2}}}$

The common factor should be canceled of $2$ .

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Cancel the common factor.

$\sqrt{- 1 + 2^{\frac{2}{2}}}$

First, divide $1$ by $1$ .

$\sqrt{- 1 + 2^{1}}$

Solve the exponent.

$\sqrt{- 1 + 2}$

Clarify by adding numbers.

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Add $- 1$ and $2$ .

$\sqrt{1}$

Any root of $1$ is $1$ .

$1$

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