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

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

Evaluate

$$14$$

Solution Steps

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

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

The square of \sqrt{3} is 3.

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

Add 3 and 4 to get 7.

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

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

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

The square of \sqrt{3} is 3.

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

Add 3 and 4 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

$$2\times 7$$