Solve {left(sqrt{33}+2sqrt{3}right)}^2

$${ \left( \sqrt{ 33 } +2 \sqrt{ 3 } \right) }^{ 2 }$$

Evaluate

$$12\sqrt{11}+45\approx 84.799497484$$

Solution Steps

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(\sqrt{33}+2\sqrt{3}\right)^{2}.

$$\left(\sqrt{33}\right)^{2}+4\sqrt{33}\sqrt{3}+4\left(\sqrt{3}\right)^{2}$$

The square of \sqrt{33} is 33.

$$33+4\sqrt{33}\sqrt{3}+4\left(\sqrt{3}\right)^{2}$$

Factor 33=3\times 11. Rewrite the square root of the product \sqrt{3\times 11} as the product of square roots \sqrt{3}\sqrt{11}.

$$33+4\sqrt{3}\sqrt{11}\sqrt{3}+4\left(\sqrt{3}\right)^{2}$$

Multiply \sqrt{3} and \sqrt{3} to get 3.

$$33+4\times 3\sqrt{11}+4\left(\sqrt{3}\right)^{2}$$

Multiply 4 and 3 to get 12.

$$33+12\sqrt{11}+4\left(\sqrt{3}\right)^{2}$$

The square of \sqrt{3} is 3.

$$33+12\sqrt{11}+4\times 3$$

Multiply 4 and 3 to get 12.

$$33+12\sqrt{11}+12$$

Add 33 and 12 to get 45.

$$45+12\sqrt{11}$$

Expand

$$12 \sqrt{11} + 45 = 84.799497484$$

Solution Steps

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(\sqrt{33}+2\sqrt{3}\right)^{2}.

$$\left(\sqrt{33}\right)^{2}+4\sqrt{33}\sqrt{3}+4\left(\sqrt{3}\right)^{2}$$

The square of \sqrt{33} is 33.

$$33+4\sqrt{33}\sqrt{3}+4\left(\sqrt{3}\right)^{2}$$

Factor 33=3\times 11. Rewrite the square root of the product \sqrt{3\times 11} as the product of square roots \sqrt{3}\sqrt{11}.

$$33+4\sqrt{3}\sqrt{11}\sqrt{3}+4\left(\sqrt{3}\right)^{2}$$

Multiply \sqrt{3} and \sqrt{3} to get 3.

$$33+4\times 3\sqrt{11}+4\left(\sqrt{3}\right)^{2}$$

Multiply 4 and 3 to get 12.

$$33+12\sqrt{11}+4\left(\sqrt{3}\right)^{2}$$

The square of \sqrt{3} is 3.

$$33+12\sqrt{11}+4\times 3$$

Multiply 4 and 3 to get 12.

$$33+12\sqrt{11}+12$$

Add 33 and 12 to get 45.

$$45+12\sqrt{11}$$