Solve (sqrt{33}+2sqrt{3})^2-4sqrt{99}

$$( \sqrt { 33 } + 2 \sqrt { 3 } ) ^ { 2 } - 4 \sqrt { 99 }$$

Evaluate

$$45$$

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}.

$$\left(\sqrt{33}\right)^{2}+4\sqrt{33}\sqrt{3}+4\left(\sqrt{3}\right)^{2}-4\sqrt{99}$$

The square of \sqrt{33} is 33.

$$33+4\sqrt{33}\sqrt{3}+4\left(\sqrt{3}\right)^{2}-4\sqrt{99}$$

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}.

$$33+4\sqrt{3}\sqrt{11}\sqrt{3}+4\left(\sqrt{3}\right)^{2}-4\sqrt{99}$$

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

$$33+4\times 3\sqrt{11}+4\left(\sqrt{3}\right)^{2}-4\sqrt{99}$$

Multiply 4 and 3 to get 12.

$$33+12\sqrt{11}+4\left(\sqrt{3}\right)^{2}-4\sqrt{99}$$

The square of \sqrt{3} is 3.

$$33+12\sqrt{11}+4\times 3-4\sqrt{99}$$

Multiply 4 and 3 to get 12.

$$33+12\sqrt{11}+12-4\sqrt{99}$$

Add 33 and 12 to get 45.

$$45+12\sqrt{11}-4\sqrt{99}$$

Factor 99=3^{2}\times 11. Rewrite the square root of the product \sqrt{3^{2}\times 11} as the product of square roots \sqrt{3^{2}}\sqrt{11}. Take the square root of 3^{2}.

$$45+12\sqrt{11}-4\times 3\sqrt{11}$$

Multiply -4 and 3 to get -12.

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

Combine 12\sqrt{11} and -12\sqrt{11} to get 0.

$$45$$

Factor

$$3^{2}\times 5$$