Solve {left(6+2sqrt{12}right)}^2=

$${ \left(6+2 \sqrt{ 12 } \right) }^{ 2 } =$$

Evaluate

$$48\sqrt{3}+84\approx 167.138438763$$

Solution Steps

Factor 12=2^{2}\times 3. Rewrite the square root of the product \sqrt{2^{2}\times 3} as the product of square roots \sqrt{2^{2}}\sqrt{3}. Take the square root of 2^{2}.

$$\left(6+2\times 2\sqrt{3}\right)^{2}$$

Multiply 2 and 2 to get 4.

$$\left(6+4\sqrt{3}\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(6+4\sqrt{3}\right)^{2}.

$$36+48\sqrt{3}+16\left(\sqrt{3}\right)^{2}$$

The square of \sqrt{3} is 3.

$$36+48\sqrt{3}+16\times 3$$

Multiply 16 and 3 to get 48.

$$36+48\sqrt{3}+48$$

Add 36 and 48 to get 84.

$$84+48\sqrt{3}$$

Expand

$$48 \sqrt{3} + 84 = 167.138438763$$

Solution Steps

Factor 12=2^{2}\times 3. Rewrite the square root of the product \sqrt{2^{2}\times 3} as the product of square roots \sqrt{2^{2}}\sqrt{3}. Take the square root of 2^{2}.

$$\left(6+2\times 2\sqrt{3}\right)^{2}$$

Multiply 2 and 2 to get 4.

$$\left(6+4\sqrt{3}\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(6+4\sqrt{3}\right)^{2}.

$$36+48\sqrt{3}+16\left(\sqrt{3}\right)^{2}$$

The square of \sqrt{3} is 3.

$$36+48\sqrt{3}+16\times 3$$

Multiply 16 and 3 to get 48.

$$36+48\sqrt{3}+48$$

Add 36 and 48 to get 84.

$$84+48\sqrt{3}$$