Solve sqrt{2^2+2^2}=sqrt{4+4}=sqrt{8}

$$\sqrt { 2 ^ { 2 } + 2 ^ { 2 } } = \sqrt { 4 + 4 } = \sqrt { 8 }$$

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Solution Steps

Calculate 2 to the power of 2 and get 4.

$$\sqrt{4+2^{2}}=\sqrt{4+4}\text{ and }\sqrt{4+4}=\sqrt{8}$$

Calculate 2 to the power of 2 and get 4.

$$\sqrt{4+4}=\sqrt{4+4}\text{ and }\sqrt{4+4}=\sqrt{8}$$

Add 4 and 4 to get 8.

$$\sqrt{8}=\sqrt{4+4}\text{ and }\sqrt{4+4}=\sqrt{8}$$

Factor 8=2^{2}\times 2. Rewrite the square root of the product \sqrt{2^{2}\times 2} as the product of square roots \sqrt{2^{2}}\sqrt{2}. Take the square root of 2^{2}.

$$2\sqrt{2}=\sqrt{4+4}\text{ and }\sqrt{4+4}=\sqrt{8}$$

Add 4 and 4 to get 8.

$$2\sqrt{2}=\sqrt{8}\text{ and }\sqrt{4+4}=\sqrt{8}$$

Factor 8=2^{2}\times 2. Rewrite the square root of the product \sqrt{2^{2}\times 2} as the product of square roots \sqrt{2^{2}}\sqrt{2}. Take the square root of 2^{2}.

$$2\sqrt{2}=2\sqrt{2}\text{ and }\sqrt{4+4}=\sqrt{8}$$

Add 4 and 4 to get 8.

$$2\sqrt{2}=2\sqrt{2}\text{ and }\sqrt{8}=\sqrt{8}$$

Factor 8=2^{2}\times 2. Rewrite the square root of the product \sqrt{2^{2}\times 2} as the product of square roots \sqrt{2^{2}}\sqrt{2}. Take the square root of 2^{2}.

$$2\sqrt{2}=2\sqrt{2}\text{ and }2\sqrt{2}=\sqrt{8}$$

Factor 8=2^{2}\times 2. Rewrite the square root of the product \sqrt{2^{2}\times 2} as the product of square roots \sqrt{2^{2}}\sqrt{2}. Take the square root of 2^{2}.

$$2\sqrt{2}=2\sqrt{2}\text{ and }2\sqrt{2}=2\sqrt{2}$$

Subtract 2\sqrt{2} from both sides.

$$2\sqrt{2}-2\sqrt{2}=0\text{ and }2\sqrt{2}=2\sqrt{2}$$

Combine 2\sqrt{2} and -2\sqrt{2} to get 0.

$$0=0\text{ and }2\sqrt{2}=2\sqrt{2}$$

Compare 0 and 0.

$$\text{true}\text{ and }2\sqrt{2}=2\sqrt{2}$$

Subtract 2\sqrt{2} from both sides.

$$\text{true}\text{ and }2\sqrt{2}-2\sqrt{2}=0$$

Combine 2\sqrt{2} and -2\sqrt{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}.

$$\text{true}$$