Solve (5sqrt{48}+sqrt{2}-6sqrt{7})divsqrt{3}

$$( 5 \sqrt { 48 } + \sqrt { 2 } - 6 \sqrt { 7 } ) \div \sqrt { 3 }$$

Evaluate

$$\frac{\sqrt{6}}{3}+20-2\sqrt{21}\approx 11.651345191$$

Solution Steps

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}.

$$\frac{5\times 4\sqrt{3}+\sqrt{2}-6\sqrt{7}}{\sqrt{3}}$$

Multiply 5 and 4 to get 20.

$$\frac{20\sqrt{3}+\sqrt{2}-6\sqrt{7}}{\sqrt{3}}$$

Rationalize the denominator of \frac{20\sqrt{3}+\sqrt{2}-6\sqrt{7}}{\sqrt{3}} by multiplying numerator and denominator by \sqrt{3}.

$$\frac{\left(20\sqrt{3}+\sqrt{2}-6\sqrt{7}\right)\sqrt{3}}{\left(\sqrt{3}\right)^{2}}$$

The square of \sqrt{3} is 3.

$$\frac{\left(20\sqrt{3}+\sqrt{2}-6\sqrt{7}\right)\sqrt{3}}{3}$$

Use the distributive property to multiply 20\sqrt{3}+\sqrt{2}-6\sqrt{7} by \sqrt{3}.

$$\frac{20\left(\sqrt{3}\right)^{2}+\sqrt{2}\sqrt{3}-6\sqrt{7}\sqrt{3}}{3}$$

The square of \sqrt{3} is 3.

$$\frac{20\times 3+\sqrt{2}\sqrt{3}-6\sqrt{7}\sqrt{3}}{3}$$

Multiply 20 and 3 to get 60.

$$\frac{60+\sqrt{2}\sqrt{3}-6\sqrt{7}\sqrt{3}}{3}$$

To multiply \sqrt{2} and \sqrt{3}, multiply the numbers under the square root.

$$\frac{60+\sqrt{6}-6\sqrt{7}\sqrt{3}}{3}$$

To multiply \sqrt{7} and \sqrt{3}, multiply the numbers under the square root.

$$\frac{60+\sqrt{6}-6\sqrt{21}}{3}$$