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

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

Evaluate

$$6\sqrt{6}+29\approx 43.696938457$$

Solution Steps

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

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

The square of \sqrt{2} is 2.

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

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

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

The square of \sqrt{3} is 3.

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

Multiply 9 and 3 to get 27.

$$2+6\sqrt{6}+27$$

Add 2 and 27 to get 29.

$$29+6\sqrt{6}$$

Expand

$$6 \sqrt{6} + 29 = 43.696938457$$

Solution Steps

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

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

The square of \sqrt{2} is 2.

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

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

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

The square of \sqrt{3} is 3.

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

Multiply 9 and 3 to get 27.

$$2+6\sqrt{6}+27$$

Add 2 and 27 to get 29.

$$29+6\sqrt{6}$$