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

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

Evaluate

$$4\sqrt{6}+11\approx 20.797958971$$

Solution Steps

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

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

The square of \sqrt{3} is 3.

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

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

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

The square of \sqrt{2} is 2.

$$3+4\sqrt{6}+4\times 2$$

Multiply 4 and 2 to get 8.

$$3+4\sqrt{6}+8$$

Add 3 and 8 to get 11.

$$11+4\sqrt{6}$$

Expand

$$4 \sqrt{6} + 11 = 20.797958971$$

Solution Steps

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

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

The square of \sqrt{3} is 3.

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

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

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

The square of \sqrt{2} is 2.

$$3+4\sqrt{6}+4\times 2$$

Multiply 4 and 2 to get 8.

$$3+4\sqrt{6}+8$$

Add 3 and 8 to get 11.

$$11+4\sqrt{6}$$