Solve {left(2+sqrt{3}right)}^2+{left(2-sqrt{3}right)}^2=

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

Evaluate

$$14$$

Solution Steps

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

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

The square of \sqrt{3} is 3.

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

Add 4 and 3 to get 7.

$$7+4\sqrt{3}+\left(2-\sqrt{3}\right)^{2}$$

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

$$7+4\sqrt{3}+4-4\sqrt{3}+\left(\sqrt{3}\right)^{2}$$

The square of \sqrt{3} is 3.

$$7+4\sqrt{3}+4-4\sqrt{3}+3$$

Add 4 and 3 to get 7.

$$7+4\sqrt{3}+7-4\sqrt{3}$$

Add 7 and 7 to get 14.

$$14+4\sqrt{3}-4\sqrt{3}$$

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

$$14$$

Factor

$$2\times 7$$