Solve 1/sqrt{2}+4sqrt{3/8}-sqrt{12}*sqrt{24}

$$\frac { 1 } { \sqrt { 2 } } + 4 \sqrt { \frac { 3 } { 8 } } - \sqrt { 12 } \times \sqrt { 24 }$$

Evaluate

$$\frac{\sqrt{2}\left(2\sqrt{3}-23\right)}{2}\approx -13.813966225$$

Short Solution Steps

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

$$\frac{\sqrt{2}}{\left(\sqrt{2}\right)^{2}}+4\sqrt{\frac{3}{8}}-\sqrt{12}\sqrt{24}$$

The square of \sqrt{2} is 2.

$$\frac{\sqrt{2}}{2}+4\sqrt{\frac{3}{8}}-\sqrt{12}\sqrt{24}$$

Rewrite the square root of the division \sqrt{\frac{3}{8}} as the division of square roots \frac{\sqrt{3}}{\sqrt{8}}.

$$\frac{\sqrt{2}}{2}+4\times \frac{\sqrt{3}}{\sqrt{8}}-\sqrt{12}\sqrt{24}$$

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}.

$$\frac{\sqrt{2}}{2}+4\times \frac{\sqrt{3}}{2\sqrt{2}}-\sqrt{12}\sqrt{24}$$

Rationalize the denominator of \frac{\sqrt{3}}{2\sqrt{2}} by multiplying numerator and denominator by \sqrt{2}.

$$\frac{\sqrt{2}}{2}+4\times \frac{\sqrt{3}\sqrt{2}}{2\left(\sqrt{2}\right)^{2}}-\sqrt{12}\sqrt{24}$$

The square of \sqrt{2} is 2.

$$\frac{\sqrt{2}}{2}+4\times \frac{\sqrt{3}\sqrt{2}}{2\times 2}-\sqrt{12}\sqrt{24}$$

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

$$\frac{\sqrt{2}}{2}+4\times \frac{\sqrt{6}}{2\times 2}-\sqrt{12}\sqrt{24}$$

Multiply 2 and 2 to get 4.

$$\frac{\sqrt{2}}{2}+4\times \frac{\sqrt{6}}{4}-\sqrt{12}\sqrt{24}$$

Cancel out 4 and 4.

$$\frac{\sqrt{2}}{2}+\sqrt{6}-\sqrt{12}\sqrt{24}$$

To add or subtract expressions, expand them to make their denominators the same. Multiply \sqrt{6} times \frac{2}{2}.

$$\frac{\sqrt{2}}{2}+\frac{2\sqrt{6}}{2}-\sqrt{12}\sqrt{24}$$

Since \frac{\sqrt{2}}{2} and \frac{2\sqrt{6}}{2} have the same denominator, add them by adding their numerators.

$$\frac{\sqrt{2}+2\sqrt{6}}{2}-\sqrt{12}\sqrt{24}$$

Factor 24=12\times 2. Rewrite the square root of the product \sqrt{12\times 2} as the product of square roots \sqrt{12}\sqrt{2}.

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

Multiply \sqrt{12} and \sqrt{12} to get 12.

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

To add or subtract expressions, expand them to make their denominators the same. Multiply -12\sqrt{2} times \frac{2}{2}.

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

Since \frac{\sqrt{2}+2\sqrt{6}}{2} and \frac{2\left(-12\right)\sqrt{2}}{2} have the same denominator, add them by adding their numerators.

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

Do the multiplications in \sqrt{2}+2\sqrt{6}+2\left(-12\right)\sqrt{2}.

$$\frac{\sqrt{2}+2\sqrt{6}-24\sqrt{2}}{2}$$

Do the calculations in \sqrt{2}+2\sqrt{6}-24\sqrt{2}.

$$\frac{-23\sqrt{2}+2\sqrt{6}}{2}$$