Solve 7sqrt{3}-5sqrt{2}divsqrt{48}+sqrt{18}

$$7 \sqrt{ 3 } -5 \sqrt{ 2 } \div \sqrt{ 48 } + \sqrt{ 18 }$$

Evaluate

$$-\frac{5\sqrt{6}}{12}+3\sqrt{2}+7\sqrt{3}\approx 15.346375614$$

Short 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}.

$$7\sqrt{3}-\frac{5\sqrt{2}}{4\sqrt{3}}+\sqrt{18}$$

Rationalize the denominator of \frac{5\sqrt{2}}{4\sqrt{3}} by multiplying numerator and denominator by \sqrt{3}.

$$7\sqrt{3}-\frac{5\sqrt{2}\sqrt{3}}{4\left(\sqrt{3}\right)^{2}}+\sqrt{18}$$

The square of \sqrt{3} is 3.

$$7\sqrt{3}-\frac{5\sqrt{2}\sqrt{3}}{4\times 3}+\sqrt{18}$$

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

$$7\sqrt{3}-\frac{5\sqrt{6}}{4\times 3}+\sqrt{18}$$

Multiply 4 and 3 to get 12.

$$7\sqrt{3}-\frac{5\sqrt{6}}{12}+\sqrt{18}$$

Factor 18=3^{2}\times 2. Rewrite the square root of the product \sqrt{3^{2}\times 2} as the product of square roots \sqrt{3^{2}}\sqrt{2}. Take the square root of 3^{2}.

$$7\sqrt{3}-\frac{5\sqrt{6}}{12}+3\sqrt{2}$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 7\sqrt{3}+3\sqrt{2} times \frac{12}{12}.

$$\frac{12\left(7\sqrt{3}+3\sqrt{2}\right)}{12}-\frac{5\sqrt{6}}{12}$$

Since \frac{12\left(7\sqrt{3}+3\sqrt{2}\right)}{12} and \frac{5\sqrt{6}}{12} have the same denominator, subtract them by subtracting their numerators.

$$\frac{12\left(7\sqrt{3}+3\sqrt{2}\right)-5\sqrt{6}}{12}$$

Do the multiplications in 12\left(7\sqrt{3}+3\sqrt{2}\right)-5\sqrt{6}.

$$\frac{84\sqrt{3}+36\sqrt{2}-5\sqrt{6}}{12}$$