Solve (sqrt{3}+sqrt{2})(sqrt{27}-sqrt{2})

$$( \sqrt { 3 } + \sqrt { 2 } ) ( \sqrt { 27 } - \sqrt { 2 } )$$

Evaluate

$$2\sqrt{6}+7\approx 11.898979486$$

Solution Steps

Factor 27=3^{2}\times 3. Rewrite the square root of the product \sqrt{3^{2}\times 3} as the product of square roots \sqrt{3^{2}}\sqrt{3}. Take the square root of 3^{2}.

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

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

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

The square of \sqrt{3} is 3.

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

Multiply 3 and 3 to get 9.

$$9-\sqrt{3}\sqrt{2}+3\sqrt{2}\sqrt{3}-\left(\sqrt{2}\right)^{2}$$

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

$$9-\sqrt{6}+3\sqrt{2}\sqrt{3}-\left(\sqrt{2}\right)^{2}$$

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

$$9-\sqrt{6}+3\sqrt{6}-\left(\sqrt{2}\right)^{2}$$

Combine -\sqrt{6} and 3\sqrt{6} to get 2\sqrt{6}.

$$9+2\sqrt{6}-\left(\sqrt{2}\right)^{2}$$

The square of \sqrt{2} is 2.

$$9+2\sqrt{6}-2$$

Subtract 2 from 9 to get 7.

$$7+2\sqrt{6}$$