Solve (sqrt{48}+sqrt{6})divsqrt{3}=

$$( \sqrt { 48 } + \sqrt { 6 } ) \div \sqrt { 3 } =$$

Evaluate

$$\sqrt{2}+4\approx 5.414213562$$

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}.

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

Rationalize the denominator of \frac{4\sqrt{3}+\sqrt{6}}{\sqrt{3}} by multiplying numerator and denominator by \sqrt{3}.

$$\frac{\left(4\sqrt{3}+\sqrt{6}\right)\sqrt{3}}{\left(\sqrt{3}\right)^{2}}$$

The square of \sqrt{3} is 3.

$$\frac{\left(4\sqrt{3}+\sqrt{6}\right)\sqrt{3}}{3}$$

Use the distributive property to multiply 4\sqrt{3}+\sqrt{6} by \sqrt{3}.

$$\frac{4\left(\sqrt{3}\right)^{2}+\sqrt{6}\sqrt{3}}{3}$$

The square of \sqrt{3} is 3.

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

Multiply 4 and 3 to get 12.

$$\frac{12+\sqrt{6}\sqrt{3}}{3}$$

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}.

$$\frac{12+\sqrt{3}\sqrt{2}\sqrt{3}}{3}$$

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

$$\frac{12+3\sqrt{2}}{3}$$

Divide each term of 12+3\sqrt{2} by 3 to get 4+\sqrt{2}.

$$4+\sqrt{2}$$