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

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

Evaluate

$$2\sqrt{6}+5\approx 9.898979486$$

Solution Steps

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

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

The square of \sqrt{3} is 3.

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

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

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

The square of \sqrt{2} is 2.

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

Add 3 and 2 to get 5.

$$5+2\sqrt{6}$$

Expand

$$2 \sqrt{6} + 5 = 9.898979486$$

Solution Steps

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

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

The square of \sqrt{3} is 3.

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

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

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

The square of \sqrt{2} is 2.

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

Add 3 and 2 to get 5.

$$5+2\sqrt{6}$$