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

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

Evaluate

$$\sqrt{6}-1\approx 1.449489743$$

Solution Steps

Apply the distributive property by multiplying each term of \sqrt{3}-\sqrt{2} by each term of \sqrt{3}+2\sqrt{2}.

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

The square of \sqrt{3} is 3.

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

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

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

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

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

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

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

The square of \sqrt{2} is 2.

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

Multiply -2 and 2 to get -4.

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

Subtract 4 from 3 to get -1.

$$-1+\sqrt{6}$$