Solve (2+sqrt{6})^2+2^2

$$( 2 + \sqrt { 6 } ) ^ { 2 } + 2 ^ { 2 }$$

Evaluate

$$4\sqrt{6}+14\approx 23.797958971$$

Solution Steps

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(2+\sqrt{6}\right)^{2}.

$$4+4\sqrt{6}+\left(\sqrt{6}\right)^{2}+2^{2}$$

The square of \sqrt{6} is 6.

$$4+4\sqrt{6}+6+2^{2}$$

Add 4 and 6 to get 10.

$$10+4\sqrt{6}+2^{2}$$

Calculate 2 to the power of 2 and get 4.

$$10+4\sqrt{6}+4$$

Add 10 and 4 to get 14.

$$14+4\sqrt{6}$$

Expand

$$4 \sqrt{6} + 14 = 23.797958971$$

Solution Steps

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(2+\sqrt{6}\right)^{2}.

$$4+4\sqrt{6}+\left(\sqrt{6}\right)^{2}+2^{2}$$

The square of \sqrt{6} is 6.

$$4+4\sqrt{6}+6+2^{2}$$

Add 4 and 6 to get 10.

$$10+4\sqrt{6}+2^{2}$$

Calculate 2 to the power of 2 and get 4.

$$10+4\sqrt{6}+4$$

Add 10 and 4 to get 14.

$$14+4\sqrt{6}$$