Solve {left(2sqrt{6}+sqrt{5}right)}^2

$${ \left(2 \sqrt{ 6 } + \sqrt{ 5 } \right) }^{ 2 }$$

Evaluate

$$4\sqrt{30}+29\approx 50.9089023$$

Solution Steps

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(2\sqrt{6}+\sqrt{5}\right)^{2}.

$$4\left(\sqrt{6}\right)^{2}+4\sqrt{6}\sqrt{5}+\left(\sqrt{5}\right)^{2}$$

The square of \sqrt{6} is 6.

$$4\times 6+4\sqrt{6}\sqrt{5}+\left(\sqrt{5}\right)^{2}$$

Multiply 4 and 6 to get 24.

$$24+4\sqrt{6}\sqrt{5}+\left(\sqrt{5}\right)^{2}$$

To multiply \sqrt{6} and \sqrt{5}, multiply the numbers under the square root.

$$24+4\sqrt{30}+\left(\sqrt{5}\right)^{2}$$

The square of \sqrt{5} is 5.

$$24+4\sqrt{30}+5$$

Add 24 and 5 to get 29.

$$29+4\sqrt{30}$$

Expand

$$4 \sqrt{30} + 29 = 50.9089023$$

Solution Steps

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(2\sqrt{6}+\sqrt{5}\right)^{2}.

$$4\left(\sqrt{6}\right)^{2}+4\sqrt{6}\sqrt{5}+\left(\sqrt{5}\right)^{2}$$

The square of \sqrt{6} is 6.

$$4\times 6+4\sqrt{6}\sqrt{5}+\left(\sqrt{5}\right)^{2}$$

Multiply 4 and 6 to get 24.

$$24+4\sqrt{6}\sqrt{5}+\left(\sqrt{5}\right)^{2}$$

To multiply \sqrt{6} and \sqrt{5}, multiply the numbers under the square root.

$$24+4\sqrt{30}+\left(\sqrt{5}\right)^{2}$$

The square of \sqrt{5} is 5.

$$24+4\sqrt{30}+5$$

Add 24 and 5 to get 29.

$$29+4\sqrt{30}$$