Solve sqrt{2}(sqrt{8}+sqrt{6})-6/sqrt{3}

$$\sqrt { 2 } ( \sqrt { 8 } + \sqrt { 6 } ) - \frac { 6 } { \sqrt { 3 } }$$

Evaluate

$$4$$

Solution Steps

Factor 8=2^{2}\times 2. Rewrite the square root of the product \sqrt{2^{2}\times 2} as the product of square roots \sqrt{2^{2}}\sqrt{2}. Take the square root of 2^{2}.

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

Rationalize the denominator of \frac{6}{\sqrt{3}} by multiplying numerator and denominator by \sqrt{3}.

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

The square of \sqrt{3} is 3.

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

Divide 6\sqrt{3} by 3 to get 2\sqrt{3}.

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

Use the distributive property to multiply \sqrt{2} by 2\sqrt{2}+\sqrt{6}.

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

The square of \sqrt{2} is 2.

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

Multiply 2 and 2 to get 4.

$$4+\sqrt{2}\sqrt{6}-2\sqrt{3}$$

Factor 6=2\times 3. Rewrite the square root of the product \sqrt{2\times 3} as the product of square roots \sqrt{2}\sqrt{3}.

$$4+\sqrt{2}\sqrt{2}\sqrt{3}-2\sqrt{3}$$

Multiply \sqrt{2} and \sqrt{2} to get 2.

$$4+2\sqrt{3}-2\sqrt{3}$$

Subtract 2\sqrt{3} from 2\sqrt{3} to get 0.

$$4$$

Factor

$$2^{2}$$