Solve frac{-sqrt{24}+sqrt{8}}{sqrt{2}}+1

$$\frac { - \sqrt { 24 } + \sqrt { 8 } } { \sqrt { 2 } } + 1$$

Evaluate

$$3-2\sqrt{3}\approx -0.464101615$$

Short Solution Steps

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}.

$$\frac{-2\sqrt{6}+\sqrt{8}}{\sqrt{2}}+1$$

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}.

$$\frac{-2\sqrt{6}+2\sqrt{2}}{\sqrt{2}}+1$$

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

$$\frac{\left(-2\sqrt{6}+2\sqrt{2}\right)\sqrt{2}}{\left(\sqrt{2}\right)^{2}}+1$$

The square of \sqrt{2} is 2.

$$\frac{\left(-2\sqrt{6}+2\sqrt{2}\right)\sqrt{2}}{2}+1$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 1 times \frac{2}{2}.

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

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

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

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

$$\frac{\left(-2\sqrt{6}\right)\sqrt{2}+2\left(\sqrt{2}\right)^{2}}{2}+1$$

The square of \sqrt{2} is 2.

$$\frac{\left(-2\sqrt{6}\right)\sqrt{2}+2\times 2}{2}+1$$

Multiply 2 and 2 to get 4.

$$\frac{\left(-2\sqrt{6}\right)\sqrt{2}+4}{2}+1$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 1 times \frac{2}{2}.

$$\frac{\left(-2\sqrt{6}\right)\sqrt{2}+4}{2}+\frac{2}{2}$$

Since \frac{\left(-2\sqrt{6}\right)\sqrt{2}+4}{2} and \frac{2}{2} have the same denominator, add them by adding their numerators.

$$\frac{\left(-2\sqrt{6}\right)\sqrt{2}+4+2}{2}$$

Multiply -1 and 2 to get -2.

$$\frac{-2\sqrt{6}\sqrt{2}+4}{2}+1$$

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}.

$$\frac{-2\sqrt{2}\sqrt{3}\sqrt{2}+4}{2}+1$$

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

$$\frac{-2\times 2\sqrt{3}+4}{2}+1$$

Multiply -2 and 2 to get -4.

$$\frac{-4\sqrt{3}+4}{2}+1$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 1 times \frac{2}{2}.

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

Since \frac{-4\sqrt{3}+4}{2} and \frac{2}{2} have the same denominator, add them by adding their numerators.

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

Do the calculations in -4\sqrt{3}+4+2.

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

Divide each term of -4\sqrt{3}+6 by 2 to get -2\sqrt{3}+3.

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