Solve sqrt{12}-sqrt{27}+frac{sqrt{6}-sqrt{2}}{sqrt{2}}

$$\sqrt { 12 } - \sqrt { 27 } + \frac { \sqrt { 6 } - \sqrt { 2 } } { \sqrt { 2 } }$$

Evaluate

$$-1$$

Short Solution Steps

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}.

$$2\sqrt{3}-\sqrt{27}+\frac{\sqrt{6}-\sqrt{2}}{\sqrt{2}}$$

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}.

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

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

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

Rationalize the denominator of \frac{\sqrt{6}-\sqrt{2}}{\sqrt{2}} by multiplying numerator and denominator by \sqrt{2}.

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

The square of \sqrt{2} is 2.

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

To add or subtract expressions, expand them to make their denominators the same. Multiply -\sqrt{3} times \frac{2}{2}.

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

Since -\frac{2\sqrt{3}}{2} and \frac{\left(\sqrt{6}-\sqrt{2}\right)\sqrt{2}}{2} have the same denominator, add them by adding their numerators.

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

Do the multiplications in -2\sqrt{3}+\left(\sqrt{6}-\sqrt{2}\right)\sqrt{2}.

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

Do the calculations in -2\sqrt{3}+2\sqrt{3}-2.

$$\frac{-2}{2}$$

Divide -2 by 2 to get -1.

$$-1$$

Factor

$$-1$$