Solve (sqrt{18}+sqrt{12})divsqrt{3}

$$( \sqrt{ 18 } + \sqrt{ 12 } ) \div \sqrt{ 3 }$$

Evaluate

$$\sqrt{6}+2\approx 4.449489743$$

Solution Steps

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}.

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

Factor 12=2^{2}\times 3. Rewrite the square root of the product \sqrt{2^{2}\times 3} as the product of square roots \sqrt{2^{2}}\sqrt{3}. Take the square root of 2^{2}.

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

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

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

The square of \sqrt{3} is 3.

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

Use the distributive property to multiply 3\sqrt{2}+2\sqrt{3} by \sqrt{3}.

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

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

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

The square of \sqrt{3} is 3.

$$\frac{3\sqrt{6}+2\times 3}{3}$$

Multiply 2 and 3 to get 6.

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

Divide each term of 3\sqrt{6}+6 by 3 to get \sqrt{6}+2.

$$\sqrt{6}+2$$