Solve frac{2sqrt{3}+3sqrt{2}}{sqrt{6}}

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

Evaluate

$$\sqrt{2}+\sqrt{3}\approx 3.14626437$$

Solution Steps

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

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

The square of \sqrt{6} is 6.

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

Use the distributive property to multiply 2\sqrt{3}+3\sqrt{2} by \sqrt{6}.

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

Factor 6=3\times 2. Rewrite the square root of the product \sqrt{3\times 2} as the product of square roots \sqrt{3}\sqrt{2}.

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

Multiply \sqrt{3} and \sqrt{3} to get 3.

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

Multiply 2 and 3 to get 6.

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

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}.

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

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

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

Multiply 3 and 2 to get 6.

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

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

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

Factor

$$\sqrt{2} + \sqrt{3} = 3.14626437$$