Question

Solve the equation ${{x}^{5}}-5{{x}^{4}}+9{{x}^{3}}-9{{x}^{2}}+5x-1=0$?

Answer 1

Hint: We start solving the problem by recalling the fact that if sum of all the coefficients in an equation is 0, then $\left( x-1 \right)$ will be one of factor of the given equation to get one of the roots of the given equation. We then find the other factor by dividing the given equation with $\left( x-1 \right)$. We then find the necessary arrangements and assume $x+\dfrac{1}{x}=y$ to proceed through the problem. We then make the necessary calculations to get the roots of y. We then equate those values to $x+\dfrac{1}{x}$ and make the necessary calculations to get the remaining roots of the given equation.

Complete step by step answer:
According to the problem, we are asked to find the given equation ${{x}^{5}}-5{{x}^{4}}+9{{x}^{3}}-9{{x}^{2}}+5x-1=0$.
Let us find the sum of all the coefficients present in the given equation.
So, we get $1-5+9-9+5-1=0$.
We know that if the sum of all the coefficients in an equation is 0, then $\left( x-1 \right)$ will be one of the factors of the given equation.
Let us divide the given equation with $\left( x-1 \right)$ as shown below:
$\begin{align}
  & 1\left| \!{\underline {\,
  \begin{array}{*{35}{l}}
   1 & -5 & +9 & -9 & +5 & -1 \\
   0 & +1 & -4 & +5 & -4 & +1 \\
\end{array} \,}} \right. \\
 & \begin{array}{*{35}{l}}
   1 & -4 & +5 & -4 & +1 & \left| \!{\underline {\,
  0 \,}} \right. \\
\end{array} \\
\end{align}$.
We can see that the quotient in the division process is ${{x}^{4}}-4{{x}^{3}}+5{{x}^{2}}-4x+1$.
So, we get $\left( x-1 \right)\left( {{x}^{4}}-4{{x}^{3}}+5{{x}^{2}}-4x+1 \right)=0$.
Now, we need to find the roots of ${{x}^{4}}-4{{x}^{3}}+5{{x}^{2}}-4x+1=0$.
So, we have ${{x}^{4}}-4{{x}^{3}}+5{{x}^{2}}-4x+1=0$.
$\Rightarrow {{x}^{2}}-4x+5-\dfrac{4}{x}+\dfrac{1}{{{x}^{2}}}=0$.
$\Rightarrow {{x}^{2}}+\dfrac{1}{{{x}^{2}}}-4x-\dfrac{4}{x}+5=0$.
$\Rightarrow {{\left( x+\dfrac{1}{x} \right)}^{2}}-2-4\left( x+\dfrac{1}{x} \right)+5=0$.
$\Rightarrow {{\left( x+\dfrac{1}{x} \right)}^{2}}-4\left( x+\dfrac{1}{x} \right)+3=0$.
Now, let us assume $x+\dfrac{1}{x}=y$.
$\Rightarrow {{y}^{2}}-4y+3=0$.
$\Rightarrow {{y}^{2}}-3y-y+3=0$.
\[\Rightarrow y\left( y-3 \right)-1\left( y-3 \right)=0\].
\[\Rightarrow \left( y-1 \right)\left( y-3 \right)=0\].
\[\Rightarrow y-1=0\], $y-3=0$.
So, we have $x+\dfrac{1}{x}=1$ and $x+\dfrac{1}{x}=3$.
$\Rightarrow \dfrac{{{x}^{2}}+1}{x}=1$, $\dfrac{{{x}^{2}}+1}{x}=3$.
$\Rightarrow {{x}^{2}}+1=x$, ${{x}^{2}}+1=3x$.
$\Rightarrow {{x}^{2}}-x+1=0$, ${{x}^{2}}-3x+1=0$.
We know that the roots of the quadratic equation $a{{x}^{2}}+bx+c=0$ is $\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$.
$\Rightarrow x=\dfrac{-\left( -1 \right)\pm \sqrt{{{\left( -1 \right)}^{2}}-4\left( 1 \right)\left( 1 \right)}}{2\left( 1 \right)}$, $x=\dfrac{-\left( -3 \right)\pm \sqrt{{{\left( -3 \right)}^{2}}-4\left( 1 \right)\left( 1 \right)}}{2\left( 1 \right)}$.
$\Rightarrow x=\dfrac{1\pm \sqrt{1-4}}{2}$, $x=\dfrac{3\pm \sqrt{9-4}}{2}$.
$\Rightarrow x=\dfrac{1\pm \sqrt{-3}}{2}$, $x=\dfrac{3\pm \sqrt{5}}{2}$.
$\Rightarrow x=\dfrac{1\pm i\sqrt{3}}{2}$, $x=\dfrac{3\pm \sqrt{5}}{2}$.

$\therefore $ We have found the roots of the given equation ${{x}^{5}}-5{{x}^{4}}+9{{x}^{3}}-9{{x}^{2}}+5x-1=0$ as 1, $\dfrac{1\pm i\sqrt{3}}{2}$, $\dfrac{3\pm \sqrt{5}}{2}$.

Note: We can see that the given problems contain huge amounts of calculations, so we need to perform each step carefully in order to avoid confusion and calculation mistakes. We can also use the fact that the quotient of the reciprocal equation will be another reciprocal equation when it is divided by $x-1$. Whenever we get this type of problem, we first try to find the sum of coefficients and sum of coefficients of odd and even terms to get one of the roots. Similarly, we can expect problems to find the roots of the equation ${{x}^{5}}+5{{x}^{4}}+9{{x}^{3}}+9{{x}^{2}}+5x+1=0$.