Question

If x = root 3 + root2/root 3 - root 2 and y = root 3 - root 2/ root 3 + root 2 then find the value of x^2 + y^2​.

Answer 1

Question:

If  $x = \dfrac{\sqrt3 +\sqrt2}{\sqrt3 - \sqrt2}$  and  $y = \dfrac{\sqrt3 - \sqrt2}{\sqrt3 + \sqrt2}$  then find the value of  $x^2+y^2.$

Solution:

We can see here that  "x  and  y  are reciprocals to each other."

$x\ =\ \dfrac{\sqrt3 +\sqrt2}{\sqrt3 - \sqrt2}\ =\ \dfrac{1}{\left(\frac{\sqrt3 - \sqrt2}{\sqrt3 + \sqrt2}\right)}=\dfrac{1}{y}\\ \\ \\ \\ y\ =\ \dfrac{\sqrt3 - \sqrt2}{\sqrt3 + \sqrt2}\ =\ \dfrac{1}{\left(\frac{\sqrt3 +\sqrt2}{\sqrt3 - \sqrt2}\right)}=\dfrac{1}{x}$

So, we remember that, the product of two numbers which are reciprocals to each other is always 1. Hence,

$\largexy=1$

Now,

Method 1:

$\begin{aligned}&x^2+y^2\\ \\ \Longrightarrow\ \ &x^2+y^2+2xy-2xy\\ \\ \Longrightarrow\ \ &(x+y)^2-2xy\\ \\ \Longrightarrow\ \ &\left(\dfrac{\sqrt3+\sqrt2}{\sqrt3-\sqrt2}+\dfrac{\sqrt3-\sqrt2}{\sqrt3+\sqrt2}\right)^2-2\times 1\\ \\ \Longrightarrow\ \ &\left(\dfrac{(\sqrt3+\sqrt2)^2+(\sqrt3-\sqrt2)^2}{(\sqrt3-\sqrt2)(\sqrt3+\sqrt2)}\right)^2-2\end{aligned}$

$\begin{aligned}\Longrightarrow\ \ &\left(\dfrac{(\sqrt3+\sqrt2)^2+(\sqrt3-\sqrt2)^2+2(\sqrt3-\sqrt2)(\sqrt3+\sqrt2)-2(\sqrt3-\sqrt2)(\sqrt3+\sqrt2)}{(\sqrt3-\sqrt2)(\sqrt3+\sqrt2)}\right)^2-2\\ \\ \Longrightarrow\ \ &\left(\dfrac{(\sqrt3+\sqrt2+\sqrt3-\sqrt2)^2-2(\sqrt3-\sqrt2)(\sqrt3+\sqrt2)}{(\sqrt3-\sqrt2)(\sqrt3+\sqrt2)}\right)^2-2\\ \\ \Longrightarrow\ \ &\left(\dfrac{(2\sqrt3)^2}{3-2}-\dfrac{2(\sqrt3-\sqrt2)(\sqrt3+\sqrt2)}{(\sqrt3-\sqrt2)(\sqrt3+\sqrt2)}\right)^2-2\end{aligned}$

$\begin{aligned}\\ \\ \Longrightarrow\ \ &(12-2)^2-2\\ \\ \Longrightarrow\ \ &10^2-2\\ \\ \Longrightarrow\ \ &100-2\\ \\ \Longrightarrow\ \ &\large\textbf{98}\end{aligned}$

Method 2:

$\begin{aligned}&x^2+y^2\\ \\ \Longrightarrow\ \ &x^2+y^2+2xy-2xy\\ \\ \Longrightarrow\ \ &(x+y)^2-2xy\\ \\ \Longrightarrow\ \ &\left(\dfrac{\sqrt3+\sqrt2}{\sqrt3-\sqrt2}+\dfrac{\sqrt3-\sqrt2}{\sqrt3+\sqrt2}\right)^2-2\times 1\\ \\ \Longrightarrow\ \ &\left(\dfrac{(\sqrt3+\sqrt2)^2+(\sqrt3-\sqrt2)^2}{(\sqrt3-\sqrt2)(\sqrt3+\sqrt2)}\right)^2-2\end{aligned}$

$\begin{aligned}\Longrightarrow\ \ &\left(\dfrac{(\sqrt3+\sqrt2)^2+(\sqrt3-\sqrt2)^2-2(\sqrt3-\sqrt2)(\sqrt3+\sqrt2)+2(\sqrt3-\sqrt2)(\sqrt3+\sqrt2)}{(\sqrt3-\sqrt2)(\sqrt3+\sqrt2)}\right)^2-2\\ \\ \Longrightarrow\ \ &\left(\dfrac{(\sqrt3+\sqrt2-\sqrt3+\sqrt2)^2+2(\sqrt3-\sqrt2)(\sqrt3+\sqrt2)}{(\sqrt3-\sqrt2)(\sqrt3+\sqrt2)}\right)^2-2\\ \\ \Longrightarrow\ \ &\left(\dfrac{(2\sqrt2)^2}{3-2}+\dfrac{2(\sqrt3-\sqrt2)(\sqrt3+\sqrt2)}{(\sqrt3-\sqrt2)(\sqrt3+\sqrt2)}\right)^2-2\end{aligned}$

$\begin{aligned}\\ \\ \Longrightarrow\ \ &(8+2)^2-2\\ \\ \Longrightarrow\ \ &10^2-2\\ \\ \Longrightarrow\ \ &100-2\\ \\ \Longrightarrow\ \ &\large\textbf{98}\end{aligned}$

Method 3:

$\begin{aligned}&x^2+y^2\\ \\ \Longrightarrow\ \ &x^2+y^2-2xy+2xy\\ \\ \Longrightarrow\ \ &(x-y)^2+2xy\\ \\ \Longrightarrow\ \ &\left(\dfrac{\sqrt3+\sqrt2}{\sqrt3-\sqrt2}-\dfrac{\sqrt3-\sqrt2}{\sqrt3+\sqrt2}\right)^2+2\times 1\\ \\ \Longrightarrow\ \ &\left(\dfrac{(\sqrt3+\sqrt2)^2-(\sqrt3-\sqrt2)^2}{(\sqrt3-\sqrt2)(\sqrt3+\sqrt2)}\right)^2+2\end{aligned}$

$\begin{aligned}\Longrightarrow\ \ &\left(\dfrac{(\sqrt3+\sqrt2+\sqrt3-\sqrt2)(\sqrt3+\sqrt2-\sqrt3+\sqrt2)}{3-2}\right)^2+2\\ \\ \Longrightarrow\ \ &\left(2\sqrt3\times2\sqrt2\right)^2+2\\ \\ \Longrightarrow\ \ &\left(4\sqrt6\right)^2+2\\ \\ \Longrightarrow\ \ &96+2\\ \\ \Longrightarrow\ \ &\large\textbf{98}\end{aligned}$

Anyways,  98  is the answer!!!