Solve sqrt{48}+frac{sqrt{6}}{4}divsqrt{27}

$$\sqrt{ 48 } + \frac{ \sqrt{ 6 } }{ 4 } \div \sqrt{ 27 }$$

Evaluate

$$\frac{\sqrt{2}}{12}+4\sqrt{3}\approx 7.04605436$$

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}.

$$4\sqrt{3}+\frac{\frac{\sqrt{6}}{4}}{\sqrt{27}}$$

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}.

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

Express \frac{\frac{\sqrt{6}}{4}}{3\sqrt{3}} as a single fraction.

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

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

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

The square of \sqrt{3} is 3.

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

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}.

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

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

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

Multiply 4 and 3 to get 12.

$$4\sqrt{3}+\frac{3\sqrt{2}}{12\times 3}$$

Multiply 12 and 3 to get 36.

$$4\sqrt{3}+\frac{3\sqrt{2}}{36}$$

Divide 3\sqrt{2} by 36 to get \frac{1}{12}\sqrt{2}.

$$4\sqrt{3}+\frac{1}{12}\sqrt{2}$$

Factor

$$\frac{\sqrt{2} + 48 \sqrt{3}}{12} = 7.046054360473267$$