Solve (sqrt{27}+sqrt{48})div2sqrt{3}

$$( \sqrt { 27 } + \sqrt { 48 } ) \div 2 \sqrt { 3 }$$

Evaluate

$$\frac{21}{2}=10.5$$

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}.

$$\frac{3\sqrt{3}+\sqrt{48}}{2}\sqrt{3}$$

Factor 48=4^{2}\times 3. Rewrite the square root of the product \sqrt{4^{2}\times 3} as the product of square roots \sqrt{4^{2}}\sqrt{3}. Take the square root of 4^{2}.

$$\frac{3\sqrt{3}+4\sqrt{3}}{2}\sqrt{3}$$

Combine 3\sqrt{3} and 4\sqrt{3} to get 7\sqrt{3}.

$$\frac{7\sqrt{3}}{2}\sqrt{3}$$

Express \frac{7\sqrt{3}}{2}\sqrt{3} as a single fraction.

$$\frac{7\sqrt{3}\sqrt{3}}{2}$$

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

$$\frac{7\times 3}{2}$$

Multiply 7 and 3 to get 21.

$$\frac{21}{2}$$

Factor

$$\frac{3 \cdot 7}{2} = 10\frac{1}{2} = 10.5$$