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

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

Evaluate

$$6\sqrt{6}+21\approx 35.696938457$$

Solution Steps

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

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

The square of \sqrt{3} is 3.

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

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

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

The square of \sqrt{2} is 2.

$$3+6\sqrt{6}+9\times 2$$

Multiply 9 and 2 to get 18.

$$3+6\sqrt{6}+18$$

Add 3 and 18 to get 21.

$$21+6\sqrt{6}$$

Expand

$$6 \sqrt{6} + 21 = 35.696938457$$

Solution Steps

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

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

The square of \sqrt{3} is 3.

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

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

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

The square of \sqrt{2} is 2.

$$3+6\sqrt{6}+9\times 2$$

Multiply 9 and 2 to get 18.

$$3+6\sqrt{6}+18$$

Add 3 and 18 to get 21.

$$21+6\sqrt{6}$$