Question

Find the value of x and y using the equations \[{{x}^{5}}-{{y}^{5}}=992\] and \[x-y=2\] .
(A) (2,4)
(B) (4,2)
(C) (4,3)
(D) (-3,-5)

Answer 1

Hint: Substitute the value of x from the equation \[x-y=2\] in \[{{x}^{5}}-{{y}^{5}}=992\]. Then, find the one root of the biquadratic equation by hit and trial method. Similarly, find the one root of the cubic equation using hit and trial method. Then find, solve the quadratic equation further and find its roots.

Complete step-by-step answer:
According to the question, we have two equations that are given
\[{{x}^{5}}-{{y}^{5}}=992\]…………(1)
\[x-y=2\]……………..(2)
From equation (2), finding the value of x in terms of y.
\[\begin{align}
  & x-y=2 \\
 & \Rightarrow x=y+2 \\
\end{align}\]
Here, we have \[x=y+2\] ………….(3)
Now, using equation (3) and substituting the value of x in equation (1), we get
\[{{(y+2)}^{5}}-{{y}^{5}}=992\]……………….(4)
We can write \[{{(y+2)}^{5}}\] as the product of \[{{(y+2)}^{3}}\] and \[{{(y+2)}^{2}}\].
Now, transforming equation (4)
\[{{(y+2)}^{5}}-{{y}^{5}}\]
\[={{(y+2)}^{3}}{{(y+2)}^{2}}-{{y}^{5}}\]…………..(5)
We know that,
\[{{(y+2)}^{2}}={{y}^{2}}+4y+4\] ……………..(6)
\[{{(y+2)}^{3}}={{y}^{3}}+6y(y+2)+8={{y}^{3}}+6{{y}^{2}}+12y+8\] ………………(7)
Now, transforming equation (5) using equation (6) and equation (7), we get
\[({{y}^{3}}+6{{y}^{2}}+12y+8)({{y}^{2}}+4y+4)-{{y}^{5}}=992\]………………(8)
Now, expanding equation (8), we get
\[\begin{align}
  & {{y}^{5}}+10{{y}^{4}}+40{{y}^{3}}+80{{y}^{2}}+80y+32-{{y}^{5}}=992 \\
 & \Rightarrow 10{{y}^{4}}+40{{y}^{3}}+80{{y}^{2}}+80y+32=992 \\
 & \Rightarrow 10{{y}^{4}}+40{{y}^{3}}+80{{y}^{2}}+80y=992-32 \\
 & \Rightarrow 10{{y}^{4}}+40{{y}^{3}}+80{{y}^{2}}+80y=960 \\
\end{align}\]
Now, dividing by 10 in the above equation, we get
\[{{y}^{4}}+4{{y}^{3}}+8{{y}^{2}}+8y-96=0\] ……………(9)
Solving by hit and trial.
Put \[(y-2)\] in the equation (9) we get,
\[16+32+32+16-96=0\]
It means \[y=2\] is the root of the equation (9).
\[(y-2)\] is a factor of the equation (9).
\[\begin{align}
  & {{y}^{4}}+4{{y}^{3}}+8{{y}^{2}}+8y-96=0 \\
 & \Rightarrow (y-2)({{y}^{3}}+6{{y}^{2}}+20y+48)=0 \\
\end{align}\]
\[\Rightarrow ({{y}^{3}}+6{{y}^{2}}+20y+48)=0\] ………………….(10)
Solving this cubic equation using a hit and trial method.
Putting \[y=-4\] in the equation(10), we get
\[-64+96-80+48=0\] .
So, \[y=-4\] is also a root of the equation (10).
It means \[(y+4)\] is the root of the cubic equation.
\[\begin{align}
  & ({{y}^{3}}+6{{y}^{2}}+20y+48)=0 \\
 & \Rightarrow (y+4)({{y}^{2}}+2y+12)=0 \\
\end{align}\]
\[\Rightarrow ({{y}^{2}}+2y+12)=0\] …………………(11)
Equation (11) is quadratic.
\[\begin{align}
  & y=\dfrac{-2\pm \sqrt{4-4(1)(12)}}{2} \\
 & \Rightarrow y=-1\pm \sqrt{-11} \\
\end{align}\]
Now, the roots are \[y=2,4,-1\pm \sqrt{-11}\] .
From equation (3), we have \[x=y+2\] .
\[x=4,-2,1\pm \sqrt{-11}\]
Hence, we have got the value of \[x=4,-2,1\pm \sqrt{-11}\] and \[y=2,4,-1\pm \sqrt{-11}\] .
So, option (B) is correct.

Note: For a quadratic equation, \[a{{x}^{2}}+bx+c=0\] .
The roots are \[\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\].
And in this question hit and trial method is also important because we are finding two roots of the biquadratic equation. So, one can make mistakes in the hit and trial method. Therefore, we should know how to apply the hit and trial method.