Question

If $x=\dfrac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}$ and $y=\dfrac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}$, find ${{x}^{2}}+{{y}^{2}}$ .

Answer 1

Hint: First we will rationalize the denominator of each x and y. Then we will substitute the obtained value of x and y in the given expression ${{x}^{2}}+{{y}^{2}}$. Then simplifying the obtained equation we will get the desired answer.

Complete step by step solution:
We have been given the values $x=\dfrac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}$ and $y=\dfrac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}$.
We have to find the value of ${{x}^{2}}+{{y}^{2}}$.
As the given values of x and y has roots in the denominator so first we will rationalize the denominators of x and y. Then we will get
$\Rightarrow x=\dfrac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}\times \dfrac{\sqrt{3}+\sqrt{2}}{\sqrt{3}+\sqrt{2}}$
Now, we know that ${{a}^{2}}+{{b}^{2}}=\left( a+b \right)\left( a-b \right)$
Now, simplifying the above obtained equation we will get
\[\begin{align}
  & \Rightarrow x=\dfrac{\left( \sqrt{3}+\sqrt{2} \right)\left( \sqrt{3}+\sqrt{2} \right)}{\left( \sqrt{3}-\sqrt{2} \right)\left( \sqrt{3}+\sqrt{2} \right)} \\
 & \Rightarrow x=\dfrac{\sqrt{3}\times \sqrt{3}+\sqrt{3}\times \sqrt{2}+\sqrt{3}\times \sqrt{2}+\sqrt{2}\times \sqrt{2}}{{{\left( \sqrt{3} \right)}^{2}}-{{\left( \sqrt{2} \right)}^{2}}} \\
 & \Rightarrow x=\dfrac{3+\sqrt{6}+\sqrt{6}+2}{3-2} \\
 & \Rightarrow x=\dfrac{5+2\sqrt{6}}{1} \\
 & \Rightarrow x=5+2\sqrt{6} \\
\end{align}\]
Now, let us rationalize the y then we will get
$\Rightarrow y=\dfrac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}\times \dfrac{\sqrt{3}-\sqrt{2}}{\sqrt{3}-\sqrt{2}}$
Now, we know that ${{a}^{2}}+{{b}^{2}}=\left( a+b \right)\left( a-b \right)$
Now, simplifying the above obtained equation we will get
\[\begin{align}
  & \Rightarrow y=\dfrac{\left( \sqrt{3}-\sqrt{2} \right)\left( \sqrt{3}-\sqrt{2} \right)}{\left( \sqrt{3}-\sqrt{2} \right)\left( \sqrt{3}+\sqrt{2} \right)} \\
 & \Rightarrow y=\dfrac{\sqrt{3}\times \sqrt{3}-\sqrt{3}\times \sqrt{2}-\sqrt{3}\times \sqrt{2}+\sqrt{2}\times \sqrt{2}}{{{\left( \sqrt{3} \right)}^{2}}-{{\left( \sqrt{2} \right)}^{2}}} \\
 & \Rightarrow y=\dfrac{3-\sqrt{6}-\sqrt{6}+2}{3-2} \\
 & \Rightarrow y=\dfrac{5-2\sqrt{6}}{1} \\
 & \Rightarrow y=5-2\sqrt{6} \\
\end{align}\]
Now, let us consider the expression ${{x}^{2}}+{{y}^{2}}$.
Now, substituting the values of x and y we will get
$= {{\left( 5+2\sqrt{6} \right)}^{2}}+{{\left( 5-2\sqrt{6} \right)}^{2}}$
Now, simplifying the above obtained equation we will get
$\begin{align}
  & = {{5}^{2}}+{{\left( 2\sqrt{6} \right)}^{2}}+2\times 5\times 2\sqrt{6}+{{5}^{2}}+{{\left( 2\sqrt{6} \right)}^{2}}-2\times 5\times 2\sqrt{6} \\
 & = 25+24+20\sqrt{6}+25+24-20\sqrt{6} \\
 & = 50+48 \\
 & = 98 \\
\end{align}$
Hence we get the value of ${{x}^{2}}+{{y}^{2}}$ as 98.

Note: Alternatively we can directly substitute the values of x and y in the given expression without rationalizing the values. The point to be remembered is that to rationalize the denominator we must multiply the numerator and denominator both by the conjugate of the denominator.