Solve 1/sqrt{2}(sqrt{6}+sqrt{12})

$$\frac { 1 } { \sqrt { 2 } } ( \sqrt { 6 } + \sqrt { 12 } )$$

Evaluate

$$\sqrt{3}+\sqrt{6}\approx 4.18154055$$

Solution Steps

Rationalize the denominator of \frac{1}{\sqrt{2}} by multiplying numerator and denominator by \sqrt{2}.

$$\frac{\sqrt{2}}{\left(\sqrt{2}\right)^{2}}\left(\sqrt{6}+\sqrt{12}\right)$$

The square of \sqrt{2} is 2.

$$\frac{\sqrt{2}}{2}\left(\sqrt{6}+\sqrt{12}\right)$$

Factor 12=2^{2}\times 3. Rewrite the square root of the product \sqrt{2^{2}\times 3} as the product of square roots \sqrt{2^{2}}\sqrt{3}. Take the square root of 2^{2}.

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

Express \frac{\sqrt{2}}{2}\left(\sqrt{6}+2\sqrt{3}\right) as a single fraction.

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

Use the distributive property to multiply \sqrt{2} by \sqrt{6}+2\sqrt{3}.

$$\frac{\sqrt{2}\sqrt{6}+2\sqrt{2}\sqrt{3}}{2}$$

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}.

$$\frac{\sqrt{2}\sqrt{2}\sqrt{3}+2\sqrt{2}\sqrt{3}}{2}$$

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

$$\frac{2\sqrt{3}+2\sqrt{2}\sqrt{3}}{2}$$

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

$$\frac{2\sqrt{3}+2\sqrt{6}}{2}$$

Divide each term of 2\sqrt{3}+2\sqrt{6} by 2 to get \sqrt{3}+\sqrt{6}.

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