Question

If we have the algebraic expression as $x-y=8$ and $xy=2$ then find the value of ${{x}^{2}}+{{y}^{2}}$.

Answer 1

HINT:- In this problem, we are going to calculate the value of one variable either $x$ or $y$ from the given equation $x-y=8$, and then we are going to substitute the value of that variable in the second given equation i.e. $xy=2$. Here we get $2$ equations. Add the two obtained equations to find the value of ${{x}^{2}}+{{y}^{2}}$

Complete step-by-step solution -
Given that $x-y=8$
The value of $x$ from the above equation is $x=y+8.....\left( \text{i} \right)$
Substituting the value of $x$ in the given equation $xy=2$ then
$\left( y+8 \right)y=2$
Use the distributive property $a\left( b+c \right)=ab+bc$ in the above equation then
\[{{y}^{2}}+8y=2\]
Now subtracting the $2$ from both sides in the above equation then
\[\begin{align}
 & {{y}^{2}}+8y-2=2-2 \\
 &\Rightarrow {{y}^{2}}+8y-2=0.......\left( \text{ii} \right)
\end{align}\]
We get the value of ${{x}^{2}}+{{y}^{2}}$ by solving equations $\left( \text{i} \right),\left( \text{ii} \right)$
Now, squaring the equation $\left( \text{i} \right)$, to find the value of $8y$ then
${{x}^{2}}={{\left( y+8 \right)}^{2}}$
We know that ${{a}^{2}}=a\times a$ then
${{x}^{2}}=\left( y+8 \right)\left( y+8 \right)$
Again, use the distributing property that $\left( a+b \right)\left( c+d \right)=\left( a+b \right)c+\left( a+b \right)d$ by treating $\left( a+b \right)$ as a single term, then
$\begin{align}
  & {{x}^{2}}=y\left( y+8 \right)+8\left( y+8 \right) \\
 &\Rightarrow {{x}^{2}}={{y}^{2}}+8y+8y+64
\end{align}$
Now subtracting the value ${{y}^{2}}+64$ on both sides then
\[\begin{align}
  & {{x}^{2}}-\left( {{y}^{2}}+64 \right)=2\left( 8y \right)+\left( {{y}^{2}}+64 \right)-\left( {{y}^{2}}+64 \right) \\
 &\Rightarrow {{x}^{2}}-{{y}^{2}}-64=2\left( 8y \right)
\end{align}\]
Now dividing the above equation with $2$ then
$\begin{align}
  & \dfrac{2\left( 8y \right)}{2}=\dfrac{{{x}^{2}}-{{y}^{2}}-64}{2} \\
 &\Rightarrow 8y=\dfrac{{{x}^{2}}-{{y}^{2}}-64}{2}
\end{align}$
Now substitute the value of $8y$ in the equation $\left( \text{ii} \right)$, then
\[\begin{align}
  & {{y}^{2}}+8y-2=0 \\
 &\Rightarrow {{y}^{2}}+\dfrac{{{x}^{2}}-{{y}^{2}}-64}{2}-2=0
\end{align}\]
Multiplying the above equation with $2$ then
\[\begin{align}
& 2{{y}^{2}}+2\left( \dfrac{{{x}^{2}}-{{y}^{2}}-64}{2} \right)-2\left( 2 \right)=2\left( 0 \right)\\
&\Rightarrow 2{{y}^{2}}+{{x}^{2}}-{{y}^{2}}-64-4=0 \\
&\Rightarrow {{x}^{2}}+{{y}^{2}}=68
\end{align}\]
Hence the value of ${{x}^{2}}+{{y}^{2}}$ is $68$.

NOTE:-We can solve this problem in another method i.e.
Squaring the given equation $x-y=8$ then
${{\left( x-y \right)}^{2}}={{8}^{2}}$
Using the formula ${{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab$ in the above equation
${{x}^{2}}+{{y}^{2}}-2xy=64$
Substitute the given value $xy=8$ in the above equation then
$\begin{align}
  & {{x}^{2}}+{{y}^{2}}-2\left( 2 \right)=64 \\
 &\Rightarrow {{x}^{2}}+{{y}^{2}}=64+4 \\
 &\Rightarrow {{x}^{2}}+{{y}^{2}}=68
\end{align}$
In both the methods we get the same answer.