Solve sqrt{27}+6/sqrt{3}=

$$\sqrt{ 27 } + \frac{ 6 }{ \sqrt{ 3 } } =$$

Evaluate

$$5\sqrt{3}\approx 8.660254038$$

Solution Steps

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}.

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

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

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

The square of \sqrt{3} is 3.

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

Divide 6\sqrt{3} by 3 to get 2\sqrt{3}.

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

Combine 3\sqrt{3} and 2\sqrt{3} to get 5\sqrt{3}.

$$5\sqrt{3}$$