Solve =frac{12sqrt{3}+24}{sqrt{2}+sqrt{6}}

$$= \frac { 12 \sqrt { 3 } + 24 } { \sqrt { 2 } + \sqrt { 6 } }$$

Evaluate

$$3\left(\sqrt{2}+\sqrt{6}\right)\approx 11.591109915$$

Solution Steps

Rationalize the denominator of \frac{12\sqrt{3}+24}{\sqrt{2}+\sqrt{6}} by multiplying numerator and denominator by \sqrt{2}-\sqrt{6}.

$$\frac{\left(12\sqrt{3}+24\right)\left(\sqrt{2}-\sqrt{6}\right)}{\left(\sqrt{2}+\sqrt{6}\right)\left(\sqrt{2}-\sqrt{6}\right)}$$

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

$$\frac{\left(12\sqrt{3}+24\right)\left(\sqrt{2}-\sqrt{6}\right)}{\left(\sqrt{2}\right)^{2}-\left(\sqrt{6}\right)^{2}}$$

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

$$\frac{\left(12\sqrt{3}+24\right)\left(\sqrt{2}-\sqrt{6}\right)}{2-6}$$

Subtract 6 from 2 to get -4.

$$\frac{\left(12\sqrt{3}+24\right)\left(\sqrt{2}-\sqrt{6}\right)}{-4}$$

Apply the distributive property by multiplying each term of 12\sqrt{3}+24 by each term of \sqrt{2}-\sqrt{6}.

$$\frac{12\sqrt{3}\sqrt{2}-12\sqrt{3}\sqrt{6}+24\sqrt{2}-24\sqrt{6}}{-4}$$

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

$$\frac{12\sqrt{6}-12\sqrt{3}\sqrt{6}+24\sqrt{2}-24\sqrt{6}}{-4}$$

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}.

$$\frac{12\sqrt{6}-12\sqrt{3}\sqrt{3}\sqrt{2}+24\sqrt{2}-24\sqrt{6}}{-4}$$

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

$$\frac{12\sqrt{6}-12\times 3\sqrt{2}+24\sqrt{2}-24\sqrt{6}}{-4}$$

Multiply -12 and 3 to get -36.

$$\frac{12\sqrt{6}-36\sqrt{2}+24\sqrt{2}-24\sqrt{6}}{-4}$$

Combine -36\sqrt{2} and 24\sqrt{2} to get -12\sqrt{2}.

$$\frac{12\sqrt{6}-12\sqrt{2}-24\sqrt{6}}{-4}$$

Combine 12\sqrt{6} and -24\sqrt{6} to get -12\sqrt{6}.

$$\frac{-12\sqrt{6}-12\sqrt{2}}{-4}$$

Factor

$$3 {(\sqrt{2} + \sqrt{6})} = 11.591109915$$