Question

If the roots of the cubic $64{{x}^{3}}-144{{x}^{2}}+92x-15=0$ are in Arithmetic Progression, then the difference between the largest and smallest root is

Answer 1

Hint: We solve this problem by first assuming the roots of the equation that are in arithmetic progression as $a-d,a,a+d$. Then we use the formula for the sum and product of the roots of the equation $a{{x}^{3}}+b{{x}^{2}}+cx+d=0$, $\alpha +\beta +\gamma =-\dfrac{b}{a}$, $\alpha \beta +\beta \gamma +\gamma \alpha =\dfrac{c}{a}$ and $\alpha \beta \gamma =-\dfrac{d}{a}$. Then we use these formulas, and substitute the assumed roots and solve them to find the value of $a$ and $d$. Then using those values, we find the roots and then the difference between the largest and the smallest roots.

Complete step-by-step solution
The equation we are given is $64{{x}^{3}}-144{{x}^{2}}+92x-15=0$.
We are also given that the roots of this cubic equation are in Arithmetic Progression. So, let us assume that the roots are $a-d,a,a+d$.
Now let us consider the formula for the sum and product of the roots of the equation, $a{{x}^{3}}+b{{x}^{2}}+cx+d=0$.
$\begin{align}
  &\Rightarrow \alpha +\beta +\gamma =-\dfrac{b}{a} \\
 &\Rightarrow \alpha \beta +\beta \gamma +\gamma \alpha =\dfrac{c}{a} \\
 &\Rightarrow \alpha \beta \gamma =-\dfrac{d}{a} \\
\end{align}$
So, using this formula, we can write the sum of the roots $a-d, a, a+d$ as,
$\begin{align}
  & \Rightarrow a-d+a+a+d=-\left( -\dfrac{144}{64} \right) \\
 & \Rightarrow 3a=\dfrac{144}{64} \\
 & \Rightarrow 3a=\dfrac{9}{4} \\
 & \Rightarrow a=\dfrac{3}{4}................\left( 1 \right) \\
\end{align}$
Now let us use the formula for the product of the roots $a-d,a,a+d$. Then we get,
\[\begin{align}
  & \Rightarrow \left( a-d \right)\times a\times \left( a+d \right)=-\left( -\dfrac{15}{64} \right) \\
 & \Rightarrow a\left( a-d \right)\left( a+d \right)=\dfrac{15}{64} \\
\end{align}\]
Now let us consider the formula,
$\left( a-b \right)\left( a+b \right)={{a}^{2}}-{{b}^{2}}$
Using this formula, we can write the above equation as,
\[\Rightarrow a\left( {{a}^{2}}-{{d}^{2}} \right)=\dfrac{15}{64}\]
Now let us substitute the value of $a$ from equation (1) in the above equation. Then we get,
\[\begin{align}
  & \Rightarrow \left( \dfrac{3}{4} \right)\left( {{\left( \dfrac{3}{4} \right)}^{2}}-{{d}^{2}} \right)=\dfrac{15}{64} \\
 & \Rightarrow \dfrac{9}{16}-{{d}^{2}}=\dfrac{\dfrac{15}{64}}{\dfrac{3}{4}} \\
 & \Rightarrow \dfrac{9}{16}-{{d}^{2}}=\dfrac{5}{16} \\
\end{align}\]
\[\begin{align}
  & \Rightarrow {{d}^{2}}=\dfrac{9}{16}-\dfrac{5}{16} \\
 & \Rightarrow {{d}^{2}}=\dfrac{4}{16} \\
 & \Rightarrow {{d}^{2}}=\dfrac{1}{4} \\
 & \Rightarrow d=\pm \dfrac{1}{2} \\
\end{align}\]
Now we need to find the difference between the largest and smallest root in the progression.
When $a=\dfrac{3}{4}\ and\ d=\dfrac{1}{2}$, the roots are
\[\begin{align}
  & \Rightarrow a-d=\dfrac{3}{4}-\dfrac{1}{2}=\dfrac{1}{4} \\
 & \Rightarrow a=\dfrac{3}{4} \\
 & \Rightarrow a+d=\dfrac{3}{4}+\dfrac{1}{2}=\dfrac{5}{4} \\
\end{align}\]
So, difference between the largest and smallest terms is,
$\Rightarrow \dfrac{5}{4}-\dfrac{1}{4}=\dfrac{4}{4}=1$
When $a=\dfrac{3}{4}\ and\ d=-\dfrac{1}{2}$, the roots are
\[\begin{align}
  & \Rightarrow a-d=\dfrac{3}{4}-\left( -\dfrac{1}{2} \right)=\dfrac{3}{4}+\dfrac{1}{2}=\dfrac{5}{4} \\
 & \Rightarrow a=\dfrac{3}{4} \\
 & \Rightarrow a+d=\dfrac{3}{4}+\left( -\dfrac{1}{2} \right)=\dfrac{3}{4}-\dfrac{1}{2}=\dfrac{1}{4} \\
\end{align}\]
So, difference between the largest and smallest roots is,
$\Rightarrow \dfrac{5}{4}-\dfrac{1}{4}=\dfrac{4}{4}=1$
So, in both the cases the difference is equal to 1. Hence the answer is 1.

Note: The common mistake one does while solving this problem is one might take the difference of largest and smallest root as,
$\begin{align}
  & \Rightarrow \left( a+d \right)-\left( a-d \right) \\
 & \Rightarrow 2d \\
\end{align}$
So, it is equal to 1 if \[d=\dfrac{1}{2}\] and -1 when $d=-\dfrac{1}{2}$.
But it is wrong because when $d=-\dfrac{1}{2}$, the largest root is $a-d$ and the smallest root is $a+d$. So, we will get the answer as 1 again.