Solve frac{sqrt{18}}{sqrt{6}+sqrt{2}}

$$\frac { \sqrt { 18 } } { \sqrt { 6 } + \sqrt { 2 } }$$

Evaluate

$$\frac{3\left(\sqrt{3}-1\right)}{2}\approx 1.098076211$$

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}.

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

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

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

Consider \left(\sqrt{6}+\sqrt{2}\right)\left(\sqrt{6}-\sqrt{2}\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

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

Square \sqrt{6}. Square \sqrt{2}.

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

Subtract 2 from 6 to get 4.

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

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

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

Factor 6=2\times 3. Rewrite the square root of the product \sqrt{2\times 3} as the product of square roots \sqrt{2}\sqrt{3}.

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

Multiply \sqrt{2} and \sqrt{2} to get 2.

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

Multiply 3 and 2 to get 6.

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

The square of \sqrt{2} is 2.

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

Multiply -3 and 2 to get -6.

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