Solve -sqrt{1/2}*sqrt{12}+sqrt{24}

$$- \sqrt { \frac { 1 } { 2 } } \times \sqrt { 12 } + \sqrt { 24 }$$

Evaluate

$$\sqrt{6}\approx 2.449489743$$

Solution Steps

Rewrite the square root of the division \sqrt{\frac{1}{2}} as the division of square roots \frac{\sqrt{1}}{\sqrt{2}}.

$$\left(-\frac{\sqrt{1}}{\sqrt{2}}\right)\sqrt{12}+\sqrt{24}$$

Calculate the square root of 1 and get 1.

$$\left(-\frac{1}{\sqrt{2}}\right)\sqrt{12}+\sqrt{24}$$

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

$$\left(-\frac{\sqrt{2}}{\left(\sqrt{2}\right)^{2}}\right)\sqrt{12}+\sqrt{24}$$

The square of \sqrt{2} is 2.

$$\left(-\frac{\sqrt{2}}{2}\right)\sqrt{12}+\sqrt{24}$$

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}.

$$\left(-\frac{\sqrt{2}}{2}\right)\times 2\sqrt{3}+\sqrt{24}$$

Cancel out 2 and 2.

$$-\sqrt{2}\sqrt{3}+\sqrt{24}$$

Factor 24=2^{2}\times 6. Rewrite the square root of the product \sqrt{2^{2}\times 6} as the product of square roots \sqrt{2^{2}}\sqrt{6}. Take the square root of 2^{2}.

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

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

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

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

$$\sqrt{6}$$