Solve sqrt{(8+sqrt{2})^2+(8-sqrt{2})^2}

$$\sqrt { ( 8 + \sqrt { 2 } ) ^ { 2 } + ( 8 - \sqrt { 2 } ) ^ { 2 } }$$

Evaluate

$$2\sqrt{33}\approx 11.489125293$$

Solution Steps

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(8+\sqrt{2}\right)^{2}.

$$\sqrt{64+16\sqrt{2}+\left(\sqrt{2}\right)^{2}+\left(8-\sqrt{2}\right)^{2}}$$

The square of \sqrt{2} is 2.

$$\sqrt{64+16\sqrt{2}+2+\left(8-\sqrt{2}\right)^{2}}$$

Add 64 and 2 to get 66.

$$\sqrt{66+16\sqrt{2}+\left(8-\sqrt{2}\right)^{2}}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(8-\sqrt{2}\right)^{2}.

$$\sqrt{66+16\sqrt{2}+64-16\sqrt{2}+\left(\sqrt{2}\right)^{2}}$$

The square of \sqrt{2} is 2.

$$\sqrt{66+16\sqrt{2}+64-16\sqrt{2}+2}$$

Add 64 and 2 to get 66.

$$\sqrt{66+16\sqrt{2}+66-16\sqrt{2}}$$

Add 66 and 66 to get 132.

$$\sqrt{132+16\sqrt{2}-16\sqrt{2}}$$

Combine 16\sqrt{2} and -16\sqrt{2} to get 0.

$$\sqrt{132}$$

Factor 132=2^{2}\times 33. Rewrite the square root of the product \sqrt{2^{2}\times 33} as the product of square roots \sqrt{2^{2}}\sqrt{33}. Take the square root of 2^{2}.

$$2\sqrt{33}$$