Solve sqrt{(2+2sqrt{3})^2+(2-2sqrt{3})^2}

$$\sqrt { ( 2 + 2 \sqrt { 3 } ) ^ { 2 } + ( 2 - 2 \sqrt { 3 } ) ^ { 2 } }$$

Evaluate

$$4\sqrt{2}\approx 5.656854249$$

Solution Steps

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(2+2\sqrt{3}\right)^{2}.

$$\sqrt{4+8\sqrt{3}+4\left(\sqrt{3}\right)^{2}+\left(2-2\sqrt{3}\right)^{2}}$$

The square of \sqrt{3} is 3.

$$\sqrt{4+8\sqrt{3}+4\times 3+\left(2-2\sqrt{3}\right)^{2}}$$

Multiply 4 and 3 to get 12.

$$\sqrt{4+8\sqrt{3}+12+\left(2-2\sqrt{3}\right)^{2}}$$

Add 4 and 12 to get 16.

$$\sqrt{16+8\sqrt{3}+\left(2-2\sqrt{3}\right)^{2}}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2-2\sqrt{3}\right)^{2}.

$$\sqrt{16+8\sqrt{3}+4-8\sqrt{3}+4\left(\sqrt{3}\right)^{2}}$$

The square of \sqrt{3} is 3.

$$\sqrt{16+8\sqrt{3}+4-8\sqrt{3}+4\times 3}$$

Multiply 4 and 3 to get 12.

$$\sqrt{16+8\sqrt{3}+4-8\sqrt{3}+12}$$

Add 4 and 12 to get 16.

$$\sqrt{16+8\sqrt{3}+16-8\sqrt{3}}$$

Add 16 and 16 to get 32.

$$\sqrt{32+8\sqrt{3}-8\sqrt{3}}$$

Combine 8\sqrt{3} and -8\sqrt{3} to get 0.

$$\sqrt{32}$$

Factor 32=4^{2}\times 2. Rewrite the square root of the product \sqrt{4^{2}\times 2} as the product of square roots \sqrt{4^{2}}\sqrt{2}. Take the square root of 4^{2}.

$$4\sqrt{2}$$