Solve {left(sqrt{5}+2right)}^2+{left(sqrt{5}-2right)}^2

$${ \left( \sqrt{ 5 } +2 \right) }^{ 2 } + { \left( \sqrt{ 5 } -2 \right) }^{ 2 }$$

Evaluate

$$18$$

Solution Steps

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(\sqrt{5}+2\right)^{2}.

$$\left(\sqrt{5}\right)^{2}+4\sqrt{5}+4+\left(\sqrt{5}-2\right)^{2}$$

The square of \sqrt{5} is 5.

$$5+4\sqrt{5}+4+\left(\sqrt{5}-2\right)^{2}$$

Add 5 and 4 to get 9.

$$9+4\sqrt{5}+\left(\sqrt{5}-2\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(\sqrt{5}-2\right)^{2}.

$$9+4\sqrt{5}+\left(\sqrt{5}\right)^{2}-4\sqrt{5}+4$$

The square of \sqrt{5} is 5.

$$9+4\sqrt{5}+5-4\sqrt{5}+4$$

Add 5 and 4 to get 9.

$$9+4\sqrt{5}+9-4\sqrt{5}$$

Add 9 and 9 to get 18.

$$18+4\sqrt{5}-4\sqrt{5}$$

Combine 4\sqrt{5} and -4\sqrt{5} to get 0.

$$18$$

Factor

$$2\times 3^{2}$$