Question

The roots of the equation ${{x}^{3}}-11{{x}^{2}}+36x-36=0$ are in harmonic progression; find the roots.

Answer 1

Hint: Here, we have to apply that if $p,q$ and $r$are the three roots of the cubic equation $a{{x}^{3}}+b{{x}^{2}}+cx+d=0$ then $p+q+r=\dfrac{-b}{a}$; $pq+qr+pr=\dfrac{c}{a}$; $pqr=\dfrac{-d}{a}$.

Complete step-by-step solution -
If the roots $p,q$ and $r$ are in HP then $\dfrac{1}{p},\dfrac{1}{q}$ and $\dfrac{1}{r}$ will be in AP.
Here, consider the equation:
${{x}^{3}}-11{{x}^{2}}+36x-36=0\text{ }.....\text{ (1)}$
Let $p,q$ and $r$ be the roots of the equation and given that the roots are in HP. Therefore we can say that $\dfrac{1}{p},\dfrac{1}{q}$ and $\dfrac{1}{r}$ will be in AP.
We also know that if $p,q$ and $r$are the three roots of the cubic equation,
 $a{{x}^{3}}+b{{x}^{2}}+cx+d=0$ …. (2) then we have:
$\begin{align}
  & p+q+r=\dfrac{-b}{a} \\
 & pq+qr+pr=\dfrac{c}{a} \\
 & pqr=\dfrac{-d}{a} \\
\end{align}$
Comparing equation (1) and equation (2) we can say that $a=1,b=-11,c=36$ and $d=-36$. Therefore we get:
$\begin{align}
  & p+q+r=11\text{ }.....\text{ (3)} \\
 & pq+qr+pr=36\text{ }......\text{ (4)} \\
 & pqr=36\text{ }......\text{ (5)} \\
\end{align}$
Now, by dividing equation (4) with equation (5) we get:
$\begin{align}
  & \dfrac{pq+pr+qr}{pqr}=\dfrac{36}{36} \\
 & \dfrac{pq}{pqr}+\dfrac{pr}{pqr}+\dfrac{qr}{pqr}=1 \\
\end{align}$
By cancellation we obtain the equation:
$\dfrac{1}{r}+\dfrac{1}{q}+\dfrac{1}{p}=1\text{ }......\text{ (6)}$
We know that $\dfrac{1}{p},\dfrac{1}{q}$ and $\dfrac{1}{r}$ are in AP. Therefore, we can say that,
$\dfrac{2}{q}=\dfrac{1}{p}+\dfrac{1}{r}\text{ }......\text{ (7)}$
By substituting equation (7) in equation (6) we get:
$\dfrac{2}{q}+\dfrac{1}{q}=1$
Next by taking the LCM we get:
$\begin{align}
  & \dfrac{2+1}{q}=1 \\
 & \dfrac{3}{q}=1 \\
 & 3=q \\
\end{align}$
Hence, we got one of the roots of equation (1) as $q=3$.
Now by putting $q=3$ in equation (3) we obtain:
$p+3+r=11$
Next, by taking 3 to the right side we get,
$\begin{align}
  & p+r=11-3 \\
 & p+r=8 \\
\end{align}$
In the next step, we substitute $q=3$ in equation (4) that yields to the equation:
$3pr=36$
Now by taking 8 to the right side we get,
$\begin{align}
  & pr=\dfrac{36}{3} \\
 & pr=12 \\
\end{align}$
Hence, we got two equations:
$\begin{align}
  & p+r=8 \\
 & pr=12 \\
\end{align}$
Next we have to solve these two equations to get $p$ and $r$.i.e.
$\begin{align}
  & p=8-r \\
 & \Rightarrow (8-r)r=12 \\
 & \Rightarrow 8r-{{r}^{2}}=12 \\
\end{align}$
By taking 12 to the left side we get a quadratic equation. i.e.
$8r-{{r}^{2}}-12=0$
By multiplying the equation by -1 we get the quadratic equation of the form:
$\begin{align}
  & {{r}^{2}}-8r+12=0 \\
 & \Rightarrow {{r}^{2}}-2r-6r+12=0 \\
 & \Rightarrow r(r-2)-6(r-2)=0 \\
\end{align}$
Since, $r-2$ is common we can take it outside. Therefore, we get:
$(r-2)(r-6)=0$
i.e $r-2=0$ or $r-6=0$
$\Rightarrow r=2$ or $r=6$
If $r=2$, we get,
$\begin{align}
  & p+2=8 \\
 & \Rightarrow p=8-2 \\
 & \Rightarrow p=6 \\
\end{align}$
If If $r=6$, we get,
$\begin{align}
  & p+6=8 \\
 & \Rightarrow p=8-6 \\
 & \Rightarrow p=2 \\
\end{align}$
Therefore, we get the other two roots as $2$ and $6$.
Hence, the three roots of the equation ${{x}^{3}}-11{{x}^{2}}+36x-36=0$ are $3,2$ and $6$

Note: An alternate method to find the roots when they are in HP is that by taking roots of the form$\dfrac{1}{a-d},\dfrac{1}{a}$ and $\dfrac{1}{a+d}$. Use the harmonic mean which represents the relation between roots and coefficients of the polynomial equation.