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

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

Evaluate

$$-4\sqrt{6}\approx -9.797958971$$

Solution Steps

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

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

The square of \sqrt{2} is 2.

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

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

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

The square of \sqrt{3} is 3.

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

Add 2 and 3 to get 5.

$$5-2\sqrt{6}-\left(\sqrt{2}+\sqrt{3}\right)^{2}$$

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

$$5-2\sqrt{6}-\left(\left(\sqrt{2}\right)^{2}+2\sqrt{2}\sqrt{3}+\left(\sqrt{3}\right)^{2}\right)$$

The square of \sqrt{2} is 2.

$$5-2\sqrt{6}-\left(2+2\sqrt{2}\sqrt{3}+\left(\sqrt{3}\right)^{2}\right)$$

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

$$5-2\sqrt{6}-\left(2+2\sqrt{6}+\left(\sqrt{3}\right)^{2}\right)$$

The square of \sqrt{3} is 3.

$$5-2\sqrt{6}-\left(2+2\sqrt{6}+3\right)$$

Add 2 and 3 to get 5.

$$5-2\sqrt{6}-\left(5+2\sqrt{6}\right)$$

To find the opposite of 5+2\sqrt{6}, find the opposite of each term.

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

Subtract 5 from 5 to get 0.

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

Combine -2\sqrt{6} and -2\sqrt{6} to get -4\sqrt{6}.

$$-4\sqrt{6}$$

Expand

$$-4\sqrt{6}$$

Solution Steps

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

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

The square of \sqrt{2} is 2.

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

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

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

The square of \sqrt{3} is 3.

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

Add 2 and 3 to get 5.

$$5-2\sqrt{6}-\left(\sqrt{2}+\sqrt{3}\right)^{2}$$

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

$$5-2\sqrt{6}-\left(\left(\sqrt{2}\right)^{2}+2\sqrt{2}\sqrt{3}+\left(\sqrt{3}\right)^{2}\right)$$

The square of \sqrt{2} is 2.

$$5-2\sqrt{6}-\left(2+2\sqrt{2}\sqrt{3}+\left(\sqrt{3}\right)^{2}\right)$$

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

$$5-2\sqrt{6}-\left(2+2\sqrt{6}+\left(\sqrt{3}\right)^{2}\right)$$

The square of \sqrt{3} is 3.

$$5-2\sqrt{6}-\left(2+2\sqrt{6}+3\right)$$

Add 2 and 3 to get 5.

$$5-2\sqrt{6}-\left(5+2\sqrt{6}\right)$$

To find the opposite of 5+2\sqrt{6}, find the opposite of each term.

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

Subtract 5 from 5 to get 0.

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

Combine -2\sqrt{6} and -2\sqrt{6} to get -4\sqrt{6}.

$$-4\sqrt{6}$$