Question

If p,q,r are in G.P and the equations, $p{{x}^{2}}+2qx+r=0$ and $d{{x}^{2}}+2ex+f=0$ have a common root, then show that $\dfrac{d}{p},\dfrac{e}{q},\dfrac{f}{r}$ are in A.P.

Answer 1

Hint: In the problem they have mentioned that $p,q,r$ are in G.P. We know that the relation of $p,q,r$ when they are in G.P i.e. ${{q}^{2}}=pr$. We will consider it as an equation. Now we will find the roots of the equation $p{{x}^{2}}+2qx+r=0$ by using the formula $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$. Here we will get the root of the equation $p{{x}^{2}}+2qx+r=0$. In the problem they also mentioned that the equations $p{{x}^{2}}+2qx+r=0$ and $d{{x}^{2}}+2ex+f=0$ have a common root. So, we will substitute the root of the equation $p{{x}^{2}}+2qx+r=0$ in the equation $d{{x}^{2}}+2ex+f=0$. Then we will have another equation and we need to simplify the obtained equation and use the relation between $p,q,r$ to show $\dfrac{d}{p},\dfrac{e}{q},\dfrac{f}{r}$ are in A.P. We know that if $a,b,c$ are A.P , then $2b=a+c$, so we need to show that $\dfrac{2e}{q}=\dfrac{d}{p}+\dfrac{f}{r}$ to say that $\dfrac{d}{p},\dfrac{e}{q},\dfrac{f}{r}$ are in A.P.

Complete step-by-step solution
Given that, $p,q,r$ are in G.P
We know that if $a,b,c$ are in G.P, then ${{b}^{2}}=ac$, hence
${{q}^{2}}=pr....\left( \text{i} \right)$
Given equation is $p{{x}^{2}}+2qx+r=0$. Comparing the above equation with $a{{x}^{2}}+bx+c=0$, then we will get
$a=p$, $b=2q$, $c=r$
We know that the root of the equation $a{{x}^{2}}+bx+c=0$ is obtained from the below formula
$x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
Hence the roots of the equation $p{{x}^{2}}+2qx+r=0$ are obtained by substituting the values $a=p$, $b=2q$, $c=r$ in the above formula then we will get,
$\begin{align}
  & x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a} \\
 & =\dfrac{-2q\pm \sqrt{{{\left( 2q \right)}^{2}}-4\left( p \right)\left( r \right)}}{2p} \\
 & =\dfrac{-2q\pm \sqrt{4{{q}^{2}}-4pr}}{2p}
\end{align}$
Taking $4$ common from the term $4{{q}^{2}}-4pr$, then we will get
$x=\dfrac{-2q\pm \sqrt{4\left( {{q}^{2}}-pr \right)}}{2p}$
From equation $\left( \text{i} \right)$ we have ${{q}^{2}}=pr$, then the value ${{q}^{2}}-pr=0$, substituting this value in the above equation, we have
$\begin{align}
  & x=\dfrac{-2q\pm \sqrt{4\left( 0 \right)}}{2p} \\
 & =\dfrac{-2q}{2p} \\
 & =\dfrac{-q}{p}
\end{align}$
Hence $-\dfrac{q}{p}$ is the root of the equation $p{{x}^{2}}+2qx+r=0$
Given that the equations $p{{x}^{2}}+2qx+r=0$ and $d{{x}^{2}}+2ex+f=0$ have same common root, then
$-\dfrac{q}{p}$ is also a root of the equation $d{{x}^{2}}+2ex+f=0$, hence substituting $x=-\dfrac{q}{p}$ in the equation $d{{x}^{2}}+2ex+f=0$, then we will get
$\begin{align}
  & d{{x}^{2}}+8ex+f=0 \\
 &\Rightarrow d{{\left( -\dfrac{q}{p} \right)}^{2}}+2e\left( -\dfrac{q}{p} \right)+f=0 \\
 &\Rightarrow \dfrac{d{{q}^{2}}}{{{p}^{2}}}-\dfrac{2eq}{p}+f=0 \\
 &\Rightarrow \dfrac{d{{q}^{2}}-2epq+f{{p}^{2}}}{{{p}^{2}}}=0 \\
 &\Rightarrow d{{q}^{2}}-2epq+f{{p}^{2}}=0.....\left( \text{ii} \right)
\end{align}$
We need to show that $\dfrac{d}{p},\dfrac{e}{q},\dfrac{f}{r}$ are in A.P.
Now dividing equation $\left( \text{ii} \right)$ with $p{{q}^{2}}$, then we will get
$\begin{align}
  &\Rightarrow d{{q}^{2}}-2epq+f{{p}^{2}}\times \left( \dfrac{1}{p{{q}^{2}}} \right)=0\times \left( \dfrac{1}{p{{q}^{2}}} \right) \\
 &\Rightarrow \dfrac{d{{q}^{2}}}{p{{q}^{2}}}-\dfrac{2epq}{p{{q}^{2}}}+\dfrac{f{{p}^{2}}}{p{{q}^{2}}}=0 \\
 &\Rightarrow \dfrac{d}{p}-\dfrac{2e}{q}+\dfrac{fp}{{{q}^{2}}}=0 \\
 &\Rightarrow \dfrac{d}{p}+\dfrac{fp}{{{q}^{2}}}=\dfrac{2e}{q}
\end{align}$
From equation $\left( \text{i} \right)$ we have ${{q}^{2}}=pr$, then we have
$\begin{align}
  &\Rightarrow \dfrac{d}{p}+\dfrac{fp}{pr}=\dfrac{2e}{q} \\
 &\Rightarrow \dfrac{d}{p}+\dfrac{f}{r}=\dfrac{2e}{q}
\end{align}$
Here we get the value of $\dfrac{d}{p}+\dfrac{f}{r}$ as $\dfrac{2e}{q}$, hence we can say that $\dfrac{d}{p},\dfrac{e}{q},\dfrac{f}{r}$ are in A.P.

Note: Students may try to find the factors of the equations $p{{x}^{2}}+2qx+r=0$ and $d{{x}^{2}}+2ex+f=0$ by factorization method. Then they may try to equate them to get the relation. But this method will be time-consuming and the chances of committing mistakes are more since the calculations might get complex.