Question

For the equation ${{x}^{3}}-7{{x}^{2}}+36=0$, if one root is twice the other, then find the other root.
A. 3
B. -6
C. 6
D. -2

Answer 1

Hint: To solve this question, we should know the formulae related to the cubic polynomial. Let us consider the equation $a{{x}^{3}}+b{{x}^{2}}+cx+d=0$. The sum of the roots of the above equation is given by $\text{sum}=\dfrac{-b}{a}$. The sum of the product of the roots taken two at a time is $\text{sum of the product of roots}=\dfrac{c}{a}$. Using these two formulas, we can write the sum and the sum of the product of the roots taken two at a time. After that by applying the condition that one root is twice the other, we get two equations in two unknowns which can be solved.

Complete step by step answer:
Let us consider the roots of the equation ${{x}^{3}}-7{{x}^{2}}+36=0$ as $p,q,r$ respectively. Let us consider that the root q is twice the root p. So, the roots become, $p,2p,r$.
Let us consider the equation $a{{x}^{3}}+b{{x}^{2}}+cx+d=0$. The sum of the roots of the above equation is given by $\text{sum}=\dfrac{-b}{a}$. The sum of the product of the roots taken two at a time is $\text{sum of the product of roots}=\dfrac{c}{a}$. In our question, we have the values of a, b, c, d as
$\begin{align}
  & a=1 \\
 & b=-7 \\
 & c=0 \\
 & d=36 \\
\end{align}$
We can write that the sum of the roots is
$\begin{align}
  & p+2p+r=\dfrac{-\left( -7 \right)}{1}=7 \\
 & 3p+r=7\to \left( 1 \right) \\
\end{align}$
We can write the sum of the product of the roots taken two at a time as
$\begin{align}
  & p\times 2p+p\times r+2p\times r=\dfrac{0}{1} \\
 & 2{{p}^{2}}+3pr=0 \\
 & p\left( 2p+3r \right)=0 \\
 & \\
\end{align}$
For two numbers a, b we know that if $ab=0$ then $a=0\text{ or }b=0$
In the above equation, if p is zero, the fundamental assumption that one root is twice the other is wrong. SO, we can write that
$\begin{align}
  & 2p+3r=0 \\
 & p=\dfrac{-3r}{2} \\
\end{align}$
Using this in equation-1, we get
$\begin{align}
  & 3\left( \dfrac{-3r}{2} \right)+r=7 \\
 & \dfrac{-9r}{2}+r=7 \\
 & r\left( 1-\dfrac{9}{2} \right)=7 \\
 & r\left( \dfrac{2-9}{2} \right)=7 \\
 & r\left( \dfrac{-7}{2} \right)=7 \\
 & r=-2 \\
\end{align}$
So, we got the third root as $r=-2$

So, the correct answer is “Option D”.

Note: In this question, students might try to solve the question using the product of the roots. If we use the two base equations as the sum and product, we will again get a cubic equation in r. That process will lead to more complex calculations and the answer might even go wrong. The trick to use in this question is that the term $2p+3r=0$ and by that observation, we can see that the calculation has decreased.