Solve 2sqrt{3}+3sqrt{2}divsqrt{3}+sqrt{2}

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

Evaluate

$$\sqrt{2}+\sqrt{6}+2\sqrt{3}\approx 7.32780492$$

Solution Steps

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

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

The square of \sqrt{3} is 3.

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

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

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

Cancel out 3 and 3.

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

Factor

$$\sqrt{2} {(\sqrt{3} + \sqrt{6} + 1)} = 7.32780492$$