Solve (6sqrt{2})^2+7^2=(6+7)^2

$$( 6 \sqrt { 2 } ) ^ { 2 } + 7 ^ { 2 } = ( 6 + 7 ) ^ { 2 }$$

Verify

Solution Steps

Expand \left(6\sqrt{2}\right)^{2}.

$$6^{2}\left(\sqrt{2}\right)^{2}+7^{2}=\left(6+7\right)^{2}$$

Calculate 6 to the power of 2 and get 36.

$$36\left(\sqrt{2}\right)^{2}+7^{2}=\left(6+7\right)^{2}$$

The square of \sqrt{2} is 2.

$$36\times 2+7^{2}=\left(6+7\right)^{2}$$

Multiply 36 and 2 to get 72.

$$72+7^{2}=\left(6+7\right)^{2}$$

Calculate 7 to the power of 2 and get 49.

$$72+49=\left(6+7\right)^{2}$$

Add 72 and 49 to get 121.

$$121=\left(6+7\right)^{2}$$

Add 6 and 7 to get 13.

$$121=13^{2}$$

Calculate 13 to the power of 2 and get 169.

$$121=169$$

Compare 121 and 169.

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