Solve frac{sqrt{24}+sqrt{8}}{sqrt{2}}-(sqrt{3})

$$\frac { \sqrt { 24 } + \sqrt { 8 } } { \sqrt { 2 } } - ( \sqrt { 3 } )$$

Evaluate

$$\sqrt{3}+2\approx 3.732050808$$

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}.

$$\frac{2\sqrt{6}+\sqrt{8}}{\sqrt{2}}-\sqrt{3}$$

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}.

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

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

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

The square of \sqrt{2} is 2.

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

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

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

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

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

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

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

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

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

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

$$\sqrt{3}+2$$