Question

Rationalize the denominator of \[\dfrac{4\sqrt{3}+5\sqrt{2}}{\sqrt{48}+\sqrt{18}}\]

Answer 1

Hint: In the first multiply and divided by the factor that is a conjugate of the denominator of the given function. Then simplify it. We will get a number that does not have an irrational number in its denominator.

Complete step-by-step answer:
First consider the given function \[\dfrac{4\sqrt{3}+5\sqrt{2}}{\sqrt{48}+\sqrt{18}}\]
It can also be written as, \[\dfrac{4\sqrt{3}+5\sqrt{2}}{4\sqrt{3}+3\sqrt{2}}\]
Next multiply and divided by the conjugate of the number \[4\sqrt{3}+3\sqrt{2}\]
That is,
             \[\begin{align}
  & =\dfrac{4\sqrt{3}+5\sqrt{2}}{4\sqrt{3}+3\sqrt{2}}\times \dfrac{4\sqrt{3}-3\sqrt{2}}{4\sqrt{3}-3\sqrt{2}} \\
 & =\dfrac{16(3)-12\sqrt{6}+20\sqrt{6}-15\sqrt{4}}{16(3)-9(2)} \\
 & =\dfrac{48-30+8\sqrt{6}}{48-18} \\
 & =\dfrac{9}{15}+\dfrac{4}{15}\sqrt{6} \\
\end{align}\]
So we got the required answer.

Note: In a situation when we have a number that contains an irrational number in its denominator. We go for a rationalization method. After rationalization, we will get a number that is independent of the irrational number in its denominator. That is what we want.