Solve sqrt{(2sqrt{3}+1)^2+(2sqrt{3}-1)^2}

$$\sqrt { ( 2 \sqrt { 3 } + 1 ) ^ { 2 } + ( 2 \sqrt { 3 } - 1 ) ^ { 2 } }$$

Evaluate

$$\sqrt{26}\approx 5.099019514$$

Solution Steps

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(2\sqrt{3}+1\right)^{2}.

$$\sqrt{4\left(\sqrt{3}\right)^{2}+4\sqrt{3}+1+\left(2\sqrt{3}-1\right)^{2}}$$

The square of \sqrt{3} is 3.

$$\sqrt{4\times 3+4\sqrt{3}+1+\left(2\sqrt{3}-1\right)^{2}}$$

Multiply 4 and 3 to get 12.

$$\sqrt{12+4\sqrt{3}+1+\left(2\sqrt{3}-1\right)^{2}}$$

Add 12 and 1 to get 13.

$$\sqrt{13+4\sqrt{3}+\left(2\sqrt{3}-1\right)^{2}}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2\sqrt{3}-1\right)^{2}.

$$\sqrt{13+4\sqrt{3}+4\left(\sqrt{3}\right)^{2}-4\sqrt{3}+1}$$

The square of \sqrt{3} is 3.

$$\sqrt{13+4\sqrt{3}+4\times 3-4\sqrt{3}+1}$$

Multiply 4 and 3 to get 12.

$$\sqrt{13+4\sqrt{3}+12-4\sqrt{3}+1}$$

Add 12 and 1 to get 13.

$$\sqrt{13+4\sqrt{3}+13-4\sqrt{3}}$$

Add 13 and 13 to get 26.

$$\sqrt{26+4\sqrt{3}-4\sqrt{3}}$$

Combine 4\sqrt{3} and -4\sqrt{3} to get 0.

$$\sqrt{26}$$