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

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

Evaluate

$$\sqrt{2}+3\approx 4.414213562$$

Solution Steps

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

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

The square of \sqrt{3} is 3.

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

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

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

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}.

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

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

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

The square of \sqrt{3} is 3.

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

Multiply 3 and 3 to get 9.

$$\frac{3\sqrt{2}+9}{3}$$

Divide each term of 3\sqrt{2}+9 by 3 to get \sqrt{2}+3.

$$\sqrt{2}+3$$