Question

Simplify: $\dfrac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}+\dfrac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}$.

Answer 1

Hint: the simplification of the given surds $\dfrac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}+\dfrac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}$ will be done in the form of addition form of taking LCM of the denominators. We apply the theorems of ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$ and ${{\left( a+b \right)}^{2}}+{{\left( a-b \right)}^{2}}=2\left( {{a}^{2}}+{{b}^{2}} \right)$to simplify the solution.

Complete step by step answer:
The given rationalisation will be solved using the LCM of the denominators.
The denominators are $\sqrt{3}-\sqrt{2}$ and $\sqrt{3}+\sqrt{2}$. We have to take their multiplication which is $\left( \sqrt{3}-\sqrt{2} \right)\left( \sqrt{3}+\sqrt{2} \right)$.
Therefore, the simplified form is
$\begin{align}
  & \dfrac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}+\dfrac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}} \\
 & =\dfrac{\left( \sqrt{3}+\sqrt{2} \right)\left( \sqrt{3}+\sqrt{2} \right)+\left( \sqrt{3}-\sqrt{2} \right)\left( \sqrt{3}-\sqrt{2} \right)}{\left( \sqrt{3}-\sqrt{2} \right)\left( \sqrt{3}+\sqrt{2} \right)} \\
 & =\dfrac{{{\left( \sqrt{3}+\sqrt{2} \right)}^{2}}+{{\left( \sqrt{3}-\sqrt{2} \right)}^{2}}}{\left( \sqrt{3}-\sqrt{2} \right)\left( \sqrt{3}+\sqrt{2} \right)} \\
\end{align}$
Now we have the identity theorem of $\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}$ which we apply for $a=\sqrt{3};b=\sqrt{2}$.
This gives $\left( \sqrt{3}-\sqrt{2} \right)\left( \sqrt{3}+\sqrt{2} \right)={{\left( \sqrt{3} \right)}^{2}}-{{\left( \sqrt{2} \right)}^{2}}=3-2=1$.
We also apply the theorem of ${{\left( a+b \right)}^{2}}+{{\left( a-b \right)}^{2}}=2\left( {{a}^{2}}+{{b}^{2}} \right)$ for $a=\sqrt{3};b=\sqrt{2}$ in the numerator and get ${{\left( \sqrt{3}+\sqrt{2} \right)}^{2}}+{{\left( \sqrt{3}-\sqrt{2} \right)}^{2}}=2\left( 3+2 \right)=10$.
Now we complete the simplification by putting the values and get
$\dfrac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}+\dfrac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}=\dfrac{{{\left( \sqrt{3}+\sqrt{2} \right)}^{2}}+{{\left( \sqrt{3}-\sqrt{2} \right)}^{2}}}{\left( \sqrt{3}-\sqrt{2} \right)\left( \sqrt{3}+\sqrt{2} \right)}=\dfrac{10}{1}=10$.

The simplified value of $\dfrac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}+\dfrac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}$ is 10.

Note: We can also use rationalisation for individual terms.
For $\dfrac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}$, we can multiply both denominator and numerator with $\sqrt{3}+\sqrt{2}$. We get
$\dfrac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}\times \dfrac{\sqrt{3}+\sqrt{2}}{\sqrt{3}+\sqrt{2}}=\dfrac{{{\left( \sqrt{3}+\sqrt{2} \right)}^{2}}}{\left( \sqrt{3}-\sqrt{2} \right)\left( \sqrt{3}+\sqrt{2} \right)}=5+2\sqrt{6}$
Doing similar thing for $\dfrac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}$, we get $\dfrac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}=5-2\sqrt{6}$.
The addition gives a solution of 10.