Solve frac{2sqrt{5}+4sqrt{3}}{sqrt{2}}-frac{2sqrt{18}-sqrt{27}}{sqrt{3}}

$$\frac { 2 \sqrt { 5 } + 4 \sqrt { 3 } } { \sqrt { 2 } } - \frac { 2 \sqrt { 18 } - \sqrt { 27 } } { \sqrt { 3 } }$$

Evaluate

$$\sqrt{10}+3\approx 6.16227766$$

Short Solution Steps

Rationalize the denominator of \frac{2\sqrt{5}+4\sqrt{3}}{\sqrt{2}} by multiplying numerator and denominator by \sqrt{2}.

$$\frac{\left(2\sqrt{5}+4\sqrt{3}\right)\sqrt{2}}{\left(\sqrt{2}\right)^{2}}-\frac{2\sqrt{18}-\sqrt{27}}{\sqrt{3}}$$

The square of \sqrt{2} is 2.

$$\frac{\left(2\sqrt{5}+4\sqrt{3}\right)\sqrt{2}}{2}-\frac{2\sqrt{18}-\sqrt{27}}{\sqrt{3}}$$

Factor 18=3^{2}\times 2. Rewrite the square root of the product \sqrt{3^{2}\times 2} as the product of square roots \sqrt{3^{2}}\sqrt{2}. Take the square root of 3^{2}.

$$\frac{\left(2\sqrt{5}+4\sqrt{3}\right)\sqrt{2}}{2}-\frac{2\times 3\sqrt{2}-\sqrt{27}}{\sqrt{3}}$$

Multiply 2 and 3 to get 6.

$$\frac{\left(2\sqrt{5}+4\sqrt{3}\right)\sqrt{2}}{2}-\frac{6\sqrt{2}-\sqrt{27}}{\sqrt{3}}$$

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}.

$$\frac{\left(2\sqrt{5}+4\sqrt{3}\right)\sqrt{2}}{2}-\frac{6\sqrt{2}-3\sqrt{3}}{\sqrt{3}}$$

Rationalize the denominator of \frac{6\sqrt{2}-3\sqrt{3}}{\sqrt{3}} by multiplying numerator and denominator by \sqrt{3}.

$$\frac{\left(2\sqrt{5}+4\sqrt{3}\right)\sqrt{2}}{2}-\frac{\left(6\sqrt{2}-3\sqrt{3}\right)\sqrt{3}}{\left(\sqrt{3}\right)^{2}}$$

The square of \sqrt{3} is 3.

$$\frac{\left(2\sqrt{5}+4\sqrt{3}\right)\sqrt{2}}{2}-\frac{\left(6\sqrt{2}-3\sqrt{3}\right)\sqrt{3}}{3}$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 2 and 3 is 6. Multiply \frac{\left(2\sqrt{5}+4\sqrt{3}\right)\sqrt{2}}{2} times \frac{3}{3}. Multiply \frac{\left(6\sqrt{2}-3\sqrt{3}\right)\sqrt{3}}{3} times \frac{2}{2}.

$$\frac{3\left(2\sqrt{5}+4\sqrt{3}\right)\sqrt{2}}{6}-\frac{2\left(6\sqrt{2}-3\sqrt{3}\right)\sqrt{3}}{6}$$

Since \frac{3\left(2\sqrt{5}+4\sqrt{3}\right)\sqrt{2}}{6} and \frac{2\left(6\sqrt{2}-3\sqrt{3}\right)\sqrt{3}}{6} have the same denominator, subtract them by subtracting their numerators.

$$\frac{3\left(2\sqrt{5}+4\sqrt{3}\right)\sqrt{2}-2\left(6\sqrt{2}-3\sqrt{3}\right)\sqrt{3}}{6}$$

Do the multiplications in 3\left(2\sqrt{5}+4\sqrt{3}\right)\sqrt{2}-2\left(6\sqrt{2}-3\sqrt{3}\right)\sqrt{3}.

$$\frac{6\sqrt{10}+12\sqrt{6}-12\sqrt{6}+18}{6}$$

Do the calculations in 6\sqrt{10}+12\sqrt{6}-12\sqrt{6}+18.

$$\frac{6\sqrt{10}+18}{6}$$

Divide each term of 6\sqrt{10}+18 by 6 to get \sqrt{10}+3.

$$\sqrt{10}+3$$