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

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

Evaluate

$$36\sqrt{2}+51\approx 101.911688245$$

Solution Steps

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

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

The square of \sqrt{6} is 6.

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

Multiply 4 and 6 to get 24.

$$24+12\sqrt{6}\sqrt{3}+9\left(\sqrt{3}\right)^{2}$$

Factor 6=3\times 2. Rewrite the square root of the product \sqrt{3\times 2} as the product of square roots \sqrt{3}\sqrt{2}.

$$24+12\sqrt{3}\sqrt{2}\sqrt{3}+9\left(\sqrt{3}\right)^{2}$$

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

$$24+12\times 3\sqrt{2}+9\left(\sqrt{3}\right)^{2}$$

Multiply 12 and 3 to get 36.

$$24+36\sqrt{2}+9\left(\sqrt{3}\right)^{2}$$

The square of \sqrt{3} is 3.

$$24+36\sqrt{2}+9\times 3$$

Multiply 9 and 3 to get 27.

$$24+36\sqrt{2}+27$$

Add 24 and 27 to get 51.

$$51+36\sqrt{2}$$

Expand

$$36 \sqrt{2} + 51 = 101.911688245$$

Solution Steps

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

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

The square of \sqrt{6} is 6.

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

Multiply 4 and 6 to get 24.

$$24+12\sqrt{6}\sqrt{3}+9\left(\sqrt{3}\right)^{2}$$

Factor 6=3\times 2. Rewrite the square root of the product \sqrt{3\times 2} as the product of square roots \sqrt{3}\sqrt{2}.

$$24+12\sqrt{3}\sqrt{2}\sqrt{3}+9\left(\sqrt{3}\right)^{2}$$

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

$$24+12\times 3\sqrt{2}+9\left(\sqrt{3}\right)^{2}$$

Multiply 12 and 3 to get 36.

$$24+36\sqrt{2}+9\left(\sqrt{3}\right)^{2}$$

The square of \sqrt{3} is 3.

$$24+36\sqrt{2}+9\times 3$$

Multiply 9 and 3 to get 27.

$$24+36\sqrt{2}+27$$

Add 24 and 27 to get 51.

$$51+36\sqrt{2}$$