Question

Find the value of x and y using the equations \[{{x}^{4}}+{{y}^{4}}=82\] and \[x-y=2\] .

Answer 1

Hint: Square the LHS and RHS of the equation \[x-y=2\] . Again, do square and then substitute \[{{x}^{4}}+{{y}^{4}}=82\]. Then, factorize the equation \[{{(xy)}^{2}}+8xy-33=0\] . Solve this equation and find the values of \[xy\] . Substitute the value of x from the equation \[x-y=2\] in the values of \[xy\]. Then, \[xy\] will become a quadratic equation in y and can be further solved.

Complete step-by-step answer:
According to the question, we have two equations that are given.
\[x-y=2\] …………….(1)
\[{{x}^{4}}+{{y}^{4}}=82\] ………………(2)
Squaring equation (1), we get
\[\begin{align}
  & {{(x-y)}^{2}}=4 \\
 & \Rightarrow {{x}^{2}}+{{y}^{2}}-2xy=4 \\
\end{align}\]
Now, taking \[-2xy\] to the RHS and squaring it, we get
\[{{({{x}^{2}}+{{y}^{2}})}^{2}}={{(4+2xy)}^{2}}\]
\[\Rightarrow {{x}^{4}}+{{y}^{4}}+2{{x}^{2}}{{y}^{2}}=16+4{{x}^{2}}{{y}^{2}}+16xy\] ……………….(3)
From equation (2), we have \[{{x}^{4}}+{{y}^{4}}=82\].
Now, using equation (2) in equation (3), we have
\[\begin{align}
  & {{x}^{4}}+{{y}^{4}}+2{{x}^{2}}{{y}^{2}}=16+4{{x}^{2}}{{y}^{2}}+16xy \\
 & \Rightarrow 82=16+2{{x}^{2}}{{y}^{2}}+16xy \\
 & \Rightarrow {{x}^{2}}{{y}^{2}}+8xy=33 \\
 & \Rightarrow {{x}^{2}}{{y}^{2}}+8xy-33=0 \\
\end{align}\]
Now, factoring the above equation, we get
\[\begin{align}
  & {{x}^{2}}{{y}^{2}}+11xy-3xy-33=0 \\
 & \Rightarrow xy(xy+11)-3(xy+11)=0 \\
 & \Rightarrow (xy-3)(xy+11)=0 \\
\end{align}\]
So, \[xy=3\] or \[xy=-11\] .
From equation (1), we have
\[x-y=2\]
\[\Rightarrow x=y+2\] ………………(4)
Taking \[xy=3\]and using equation (4), we have
\[\begin{align}
  & y(y+2)=3 \\
 & \Rightarrow {{y}^{2}}+2y-3=0 \\
\end{align}\]
Factorising the above quadratic equation,
\[\begin{align}
  & {{y}^{2}}+3y-y-3=0 \\
 & \Rightarrow y(y+3)-1(y+3)=0 \\
 & \Rightarrow (y-1)(y+3)=0 \\
\end{align}\]
\[y=1\] or \[y=-3\] .
Now, taking \[xy=-11\] and using equation (4), we have
\[\begin{align}
  & y(y+2)=-11 \\
 & \Rightarrow {{y}^{2}}+2y+11=0 \\
\end{align}\]
Discriminant \[=\sqrt{4-44}\]
Discriminant < 0.
So, there are no real values of y. Therefore, we cannot take \[xy=-11\] .
Therefore, \[y=1\] or \[y=-3\] and \[x=3\] or \[x=-1\].

Note: In this question, one can take \[xy=-11\] and then solve the quadratic equation. But that quadratic equation has discriminant less than zero. So, real values of y will not exist. Only, the imaginary values exist. We have to find the solution to the question. Therefore, we only need real values of x and y.