Solve frac{sqrt{96}}{3}+sqrt{51}=

$$\frac { \sqrt { 96 } } { 3 } + \sqrt { 51 } =$$

Evaluate

$$\frac{4\sqrt{6}}{3}+\sqrt{51}\approx 10.407414752$$

Short Solution Steps

Factor 96=4^{2}\times 6. Rewrite the square root of the product \sqrt{4^{2}\times 6} as the product of square roots \sqrt{4^{2}}\sqrt{6}. Take the square root of 4^{2}.

$$\frac{4\sqrt{6}}{3}+\sqrt{51}$$

To add or subtract expressions, expand them to make their denominators the same. Multiply \sqrt{51} times \frac{3}{3}.

$$\frac{4\sqrt{6}}{3}+\frac{3\sqrt{51}}{3}$$

Since \frac{4\sqrt{6}}{3} and \frac{3\sqrt{51}}{3} have the same denominator, add them by adding their numerators.

$$\frac{4\sqrt{6}+3\sqrt{51}}{3}$$

Factor

$$\frac{3 \sqrt{51} + 4 \sqrt{6}}{3} = 10.407414752253754$$