Question

If \[a=\dfrac{4\sqrt{6}}{\sqrt{2}+\sqrt{3}}\], then find the value of:
\[\dfrac{a+2\sqrt{2}}{a-2\sqrt{2}}+\dfrac{a+2\sqrt{3}}{a-2\sqrt{3}}\]

Answer 1

Hint: First of all we will rationalize the term of ‘a’ by multiplying the conjugate of denominator of term ‘a’ to numerator as well as denominator i.e. \[\sqrt{2}-\sqrt{3}\]. After rationalization, we will substitute the value of ‘a’ in the expression to find the required value.

Complete step-by-step answer:

We have been given \[a=\dfrac{4\sqrt{6}}{\sqrt{2}+\sqrt{3}}\] and asked to find the value of \[\dfrac{a+2\sqrt{2}}{a-2\sqrt{2}}+\dfrac{a+2\sqrt{3}}{a-2\sqrt{3}}\].
First of all we will rationalize the term ‘a’ by multiplying its numerator as well as denominator by the conjugate of its denominator.
We know that the irrational conjugate of an irrational number \[x+\sqrt{y}\] is \[\left( x-\sqrt{y} \right)\].
We have,
\[a=\dfrac{4\sqrt{6}}{\sqrt{2}+\sqrt{3}}=\dfrac{4\sqrt{6}}{\sqrt{3}+\sqrt{2}}\]
Now, the conjugate of \[\left( \sqrt{3}+\sqrt{2} \right)\] is \[\left( \sqrt{3}-\sqrt{2} \right)\].
So we will multiply the numerator as well as the denominator by its conjugate.
Since we know the identity, \[\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}\]
\[\begin{align}
  & \Rightarrow a=\dfrac{4\sqrt{6}\times \left( \sqrt{3}-\sqrt{2} \right)}{\left( \sqrt{3}+\sqrt{2} \right)\times \left( \sqrt{3}-\sqrt{2} \right)}=\dfrac{4\sqrt{18}-4\sqrt{12}}{{{\left( \sqrt{3} \right)}^{2}}-{{\left( \sqrt{2} \right)}^{2}}} \\
 & \Rightarrow a=\dfrac{4\sqrt{18}-4\sqrt{12}}{3-2}=\dfrac{4\sqrt{{{3}^{2}}\times 2}-4\sqrt{{{2}^{2}}\times 3}}{1} \\
 & \Rightarrow a=12\sqrt{2}-8\sqrt{3} \\
\end{align}\]
We have been asked to find the value of \[\dfrac{a+2\sqrt{2}}{a-2\sqrt{2}}+\dfrac{a+2\sqrt{3}}{a-2\sqrt{3}}\].
Now substituting the value of \[a=12\sqrt{2}-8\sqrt{3}\], which we get after rationalizing, in the above expression, we get as follows:
\[\begin{align}
  & \Rightarrow \dfrac{12\sqrt{2}-8\sqrt{3}+2\sqrt{2}}{12\sqrt{2}-8\sqrt{3}-2\sqrt{2}}+\dfrac{12\sqrt{2}-8\sqrt{3}+2\sqrt{3}}{12\sqrt{2}-8\sqrt{3}-2\sqrt{3}} \\
 & =\dfrac{14\sqrt{2}-8\sqrt{3}}{10\sqrt{2}-8\sqrt{3}}+\dfrac{12\sqrt{2}-6\sqrt{3}}{12\sqrt{2}-10\sqrt{3}} \\
 & =\dfrac{2\left( 7\sqrt{2}-4\sqrt{3} \right)}{2\left( 5\sqrt{2}-4\sqrt{3} \right)}+\dfrac{6\left( 2\sqrt{2}-\sqrt{3} \right)}{2\left( 6\sqrt{2}-5\sqrt{3} \right)} \\
 & =\dfrac{7\sqrt{2}-4\sqrt{3}}{5\sqrt{2}-4\sqrt{3}}+\dfrac{3\left( 2\sqrt{2}-\sqrt{3} \right)}{6\sqrt{2}-5\sqrt{3}} \\
\end{align}\]
Rationalizing each term, we get as follows:
\[\Rightarrow \dfrac{7\sqrt{2}-4\sqrt{3}\times \left( 5\sqrt{2}+4\sqrt{3} \right)}{\left( 5\sqrt{2}-4\sqrt{3} \right)\left( 5\sqrt{2}+4\sqrt{3} \right)}+\left[ \dfrac{3\left( 2\sqrt{2}-\sqrt{3} \right)\times \left( 6\sqrt{2}+5\sqrt{3} \right)}{\left( 6\sqrt{2}-5\sqrt{3} \right)\left( 6\sqrt{2}+5\sqrt{3} \right)} \right]\]
Since we know the identity, \[\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}\]
\[\begin{align}
  & \Rightarrow \dfrac{35{{\left( \sqrt{2} \right)}^{2}}+28\sqrt{6}-20\sqrt{6}-16{{\left( \sqrt{3} \right)}^{2}}}{{{\left( 5\sqrt{2} \right)}^{2}}-{{\left( 4\sqrt{3} \right)}^{2}}}+3\left[ \dfrac{12{{\left( \sqrt{2} \right)}^{2}}+10\sqrt{6}-6\sqrt{6}-5{{\left( \sqrt{3} \right)}^{2}}}{{{\left( 6\sqrt{2} \right)}^{2}}-{{\left( 5\sqrt{3} \right)}^{2}}} \right] \\
 & =\dfrac{70+8\sqrt{6}-48}{50-48}+3\left[ \dfrac{24+4\sqrt{6}-15}{72-75} \right] \\
 & =\dfrac{22+8\sqrt{6}}{2}+3\left[ \dfrac{9+4\sqrt{6}}{-3} \right] \\
 & =\dfrac{2\left( 11+4\sqrt{6} \right)}{2}-3\left[ \dfrac{9+4\sqrt{6}}{3} \right] \\
 & =11+4\sqrt{6}-9-4\sqrt{6} \\
\end{align}\]
Cancelling similar terms, we get as follows:
\[\Rightarrow 11-9=2\]
Therefore, the value of the given expression is equal to 2.

Note: Be careful while doing calculation and also take care of the sign after rationalization because it is a chance that you might make a sign mistake while doing rationalization. Remember that to rationalize, the term \[\dfrac{x+\sqrt{y}}{x-\sqrt{y}}\] we multiply the numerator as well as denominator of the term by the conjugate of the denominator \[\left( x-\sqrt{y} \right)\] which is equal to same but the sign is opposite of it, i.e. \[\left( x+\sqrt{y} \right)\].