Solve sqrt{(sqrt{7}+1)^2+(sqrt{7}-1)^2}

$$\sqrt { ( \sqrt { 7 } + 1 ) ^ { 2 } + ( \sqrt { 7 } - 1 ) ^ { 2 } }$$

Evaluate

$$4$$

Solution Steps

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(\sqrt{7}+1\right)^{2}.

$$\sqrt{\left(\sqrt{7}\right)^{2}+2\sqrt{7}+1+\left(\sqrt{7}-1\right)^{2}}$$

The square of \sqrt{7} is 7.

$$\sqrt{7+2\sqrt{7}+1+\left(\sqrt{7}-1\right)^{2}}$$

Add 7 and 1 to get 8.

$$\sqrt{8+2\sqrt{7}+\left(\sqrt{7}-1\right)^{2}}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(\sqrt{7}-1\right)^{2}.

$$\sqrt{8+2\sqrt{7}+\left(\sqrt{7}\right)^{2}-2\sqrt{7}+1}$$

The square of \sqrt{7} is 7.

$$\sqrt{8+2\sqrt{7}+7-2\sqrt{7}+1}$$

Add 7 and 1 to get 8.

$$\sqrt{8+2\sqrt{7}+8-2\sqrt{7}}$$

Add 8 and 8 to get 16.

$$\sqrt{16+2\sqrt{7}-2\sqrt{7}}$$

Combine 2\sqrt{7} and -2\sqrt{7} to get 0.

$$\sqrt{16}$$

Calculate the square root of 16 and get 4.

$$4$$

Factor

$$2^{2}$$