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

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

Evaluate

$$5-2\sqrt{6}\approx 0.101020514$$

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

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

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.

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