Solve 12sqrt{6}+(3sqrt{3}+2sqrt{2})^2+36

$$12 \sqrt { 6 } + ( 3 \sqrt { 3 } + 2 \sqrt { 2 } ) ^ { 2 } + 36$$

Evaluate

$$24\sqrt{6}+71\approx 129.787753827$$

Solution Steps

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

$$12\sqrt{6}+9\left(\sqrt{3}\right)^{2}+12\sqrt{3}\sqrt{2}+4\left(\sqrt{2}\right)^{2}+36$$

The square of \sqrt{3} is 3.

$$12\sqrt{6}+9\times 3+12\sqrt{3}\sqrt{2}+4\left(\sqrt{2}\right)^{2}+36$$

Multiply 9 and 3 to get 27.

$$12\sqrt{6}+27+12\sqrt{3}\sqrt{2}+4\left(\sqrt{2}\right)^{2}+36$$

To multiply \sqrt{3} and \sqrt{2}, multiply the numbers under the square root.

$$12\sqrt{6}+27+12\sqrt{6}+4\left(\sqrt{2}\right)^{2}+36$$

The square of \sqrt{2} is 2.

$$12\sqrt{6}+27+12\sqrt{6}+4\times 2+36$$

Multiply 4 and 2 to get 8.

$$12\sqrt{6}+27+12\sqrt{6}+8+36$$

Add 27 and 8 to get 35.

$$12\sqrt{6}+35+12\sqrt{6}+36$$

Combine 12\sqrt{6} and 12\sqrt{6} to get 24\sqrt{6}.

$$24\sqrt{6}+35+36$$

Add 35 and 36 to get 71.

$$24\sqrt{6}+71$$