Question

Solve the reciprocal equation:
$6{{x}^{4}}-37{{x}^{3}}+12{{x}^{2}}-37x+6=0$

Answer 1

Hint: Here, we will first divide the equation by $6{{x}^{2}}$ and then we will replace $\left( x+\dfrac{1}{x} \right)$ by y to obtain a quadratic equation in terms of y which in turn help in obtaining the values of x.

Complete step-by-step answer: A polynomial P(x) of degree n is said to be a reciprocal polynomial if one of the following conditions is satisfied:
(1) $P\left( x \right)={{x}^{n}}P\left( \dfrac{1}{x} \right)$
(2) $P\left( x \right)=-{{x}^{n}}P\left( \dfrac{1}{x} \right)$
Since, the equation given here is $6{{x}^{4}}-37{{x}^{3}}+12{{x}^{2}}-37x+6=0$
On dividing by $6{{x}^{2}}$ on both sides of the equation, we get:
$\dfrac{6{{x}^{4}}}{6{{x}^{2}}}-\dfrac{37{{x}^{3}}}{6{{x}^{2}}}+\dfrac{12{{x}^{2}}}{6{{x}^{2}}}-\dfrac{37x}{6{{x}^{2}}}+\dfrac{6}{6{{x}^{2}}}=0$
$\Rightarrow {{x}^{2}}-\dfrac{37}{6}+2-\dfrac{37}{6x}+\dfrac{1}{{{x}^{2}}}=0$
On simplifying, we get:
$\begin{align}
  & {{x}^{2}}+\dfrac{1}{{{x}^{2}}}-\dfrac{37x}{6}-\dfrac{37}{6x}+2=0 \\
 & \Rightarrow \left( {{x}^{2}}+\dfrac{1}{{{x}^{2}}} \right)-\dfrac{37}{6}\left( x+\dfrac{1}{x} \right)+2=0.............\left( 1 \right) \\
\end{align}$
Now, we will put $\left( x+\dfrac{1}{x} \right)=y$
On squaring it both sides, we get:
$\begin{align}
  & {{\left( x+\dfrac{1}{x} \right)}^{2}}={{y}^{2}} \\
 & \Rightarrow {{x}^{2}}+\dfrac{1}{{{x}^{2}}}+2\times x\times \dfrac{1}{x}={{y}^{2}} \\
 & \Rightarrow \left( {{x}^{2}}+\dfrac{1}{{{x}^{2}}} \right)=\left( {{y}^{2}}-2 \right) \\
\end{align}$
Now, on substituting these values in equation (1), we get:
 ${{y}^{2}}-2-\dfrac{37}{6}y+2=0$
On multiplying by 6 on both sides of this equation, we get:
$\begin{align}
  & 6{{y}^{2}}-12-37y+12=0 \\
 & \Rightarrow 6{{y}^{2}}-37y=0 \\
\end{align}$
On taking y as common, we get:
$y\left( 6y-37 \right)=0$
So, we have:
$y=0$ or $6y-37=0\Rightarrow y=\dfrac{37}{6}$
From this, we get that:
$x+\dfrac{1}{x}=0$ or $x+\dfrac{1}{x}=\dfrac{37}{6}$
Now, when $x+\dfrac{1}{x}=0$, we have:
$\begin{align}
  & \dfrac{{{x}^{2}}+1}{x}=0 \\
 & \Rightarrow {{x}^{2}}+1=0 \\
 & \Rightarrow x=\pm \sqrt{-1}=\pm i \\
\end{align}$
And, when $x+\dfrac{1}{x}=\dfrac{37}{6}$, we have:
$\begin{align}
  & 6\left( {{x}^{2}}+1 \right)=37x \\
 & \Rightarrow 6{{x}^{2}}-37x+6=0 \\
\end{align}$
We know that the solution for a general quadratic equation of the form $a{{x}^{2}}+bx+c=0$ is given as:
$x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ .
So, using this formula the solution of the quadratic equation $6{{x}^{2}}-37x+6=0$ is given as:
$\begin{align}
  & x=\dfrac{-\left( -37 \right)\pm \sqrt{{{\left( -37 \right)}^{2}}-4\times 6\times 6}}{2\times 6} \\
 & \Rightarrow x=\dfrac{-\left( -37 \right)\pm \sqrt{1369-144}}{12} \\
 & \Rightarrow x=\dfrac{37\pm \sqrt{1225}}{12}=\dfrac{37\pm 35}{12} \\
\end{align}$
So, $x=\dfrac{37+35}{12}=\dfrac{72}{12}=6$ or $x=\dfrac{37-35}{12}=\dfrac{2}{12}=\dfrac{1}{6}$
So, the two values of x from this equation come out to be x = 12 or x=6.
Hence, the values of x are $i,-i,6\text{ and }\dfrac{1}{6}$.

Note: Students should note here that since the degree of the given equation is 4, that is why we get 4 roots of the equation or 4 values of x. Always remember that solution for a general quadratic equation of the form \[a{{x}^{2}}+bx+c=0\] is given as $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ also remember that ${{(a+b)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$. Also, remember that $\sqrt{-1}$ is a complex number and is denoted by i. The calculations must be done properly to avoid mistakes as if calculation went wrong is the starting of a solution then further solving gets more complex and it will be hard to solve the question.