Question

Given that \[\alpha ,\gamma \] are the roots of the equation \[A{{x}^{2}}-4x+1=0\] and \[\beta ,\delta \] are the roots of the equation \[B{{x}^{2}}-6x+1=0\], the values of\[A\] and \[B\] such that \[\alpha ,\beta ,\gamma \] and $\delta $ are in H.P. are
A. \[A=3, B=8\]
B. \[A=-3, B=8\]
C. \[A=3, B=-8\]
D. None of these

Answer 1

Hint: In this question, we have to find the values of the variables \[A\] and \[B\]. These values are obtained from the given quadratic expressions by using the given roots \[\alpha,\beta,\gamma \] and $\delta $ which are in H.P. For this, we need to calculate the sum of roots and the product of roots. With these values, we can able to find the required variables.

Formula Used: The relationship between A.P and H.P:
If $a,b,c$ are in A.P then $\dfrac{1}{a},\dfrac{1}{b},\dfrac{1}{c}$ are in H.P.
Or
If $a,b,c$ are H.P then $\dfrac{1}{a},\dfrac{1}{b},\dfrac{1}{c}$ are in H.P.
The general terms in an arithmetic series are $a,a+d,a+2d,...,a+nd$ where the common difference $d={{t}_{n}}-{{t}_{n-1}}$.
If $\alpha ,\beta $ are the roots of a quadratic equation $a{{x}^{2}}+bx+c=0$, then the sum and product of the roots are
$\begin{align}
  & \alpha +\beta =\dfrac{-b}{a} \\
 & \alpha \cdot \beta =\dfrac{c}{a} \\
\end{align}$

Complete step by step solution: It is given that, \[\alpha ,\beta ,\gamma ,\delta \] are in H.P. So, \[\dfrac{1}{\alpha },\dfrac{1}{\beta },\dfrac{1}{\gamma },\dfrac{1}{\delta }\] are in A.P.
Then, the terms of the series are on comparing with the general terms of A.P, and we get
\[\dfrac{1}{\alpha }=a;\dfrac{1}{\beta }=\dfrac{1}{\alpha }+d;\dfrac{1}{\gamma }=\dfrac{1}{\alpha }+2d;\dfrac{1}{\delta }=\dfrac{1}{\alpha }+3d\]
It is given that, \[\alpha,\gamma \] are the roots of the equation \[A{{x}^{2}}-4x+1=0\].
Then, their sum and products are
\[\begin{align}
  & \alpha +\gamma =\dfrac{-(-4)}{A}=\dfrac{4}{A} \\
 & \alpha \cdot \gamma =\dfrac{1}{A} \\
\end{align}\]
On dividing the sum and the products, we get
\[\begin{align}
  & \dfrac{\alpha +\gamma }{\alpha \cdot \gamma }=\dfrac{{}^{4}/{}_{A}}{{}^{1}/{}_{A}}=4 \\
 & \Rightarrow \dfrac{1}{\alpha }+\dfrac{1}{\gamma }=4 \\
\end{align}\]
But we know \[\dfrac{1}{\gamma }=\dfrac{1}{\alpha }+2d\]
Then, on substituting, we get
\[\begin{align}
  & \Rightarrow \dfrac{1}{\alpha }+\dfrac{1}{\alpha }+2d=4 \\
 & \Rightarrow \dfrac{2}{\alpha }+2d=4 \\
 & \Rightarrow \dfrac{1}{\alpha }+d=2\text{ }...(1) \\
\end{align}\]
It is given that, \[\beta ,\delta \] are the roots of the equation \[B{{x}^{2}}-6x+1=0\].
Then, their sum and products are
\[\begin{align}
  & \beta +\delta =\dfrac{-(-6)}{B}=\dfrac{6}{B} \\
 & \beta \cdot \delta =\dfrac{1}{B} \\
\end{align}\]
On dividing the sum and the products, we get
\[\begin{align}
  & \dfrac{\beta +\delta }{\beta \cdot \delta }=\dfrac{{}^{6}/{}_{B}}{{}^{1}/{}_{B}}=6 \\
 & \Rightarrow \dfrac{1}{\beta }+\dfrac{1}{\delta }=6 \\
\end{align}\]
But we know \[\dfrac{1}{\beta }=\dfrac{1}{\alpha }+d;\dfrac{1}{\delta }=\dfrac{1}{\alpha }+3d\]
Then, on substituting, we get
\[\begin{align}
  & \dfrac{1}{\alpha }+d+\dfrac{1}{\alpha }+3d=6 \\
 & \dfrac{2}{\alpha }+4d=6 \\
 & \dfrac{1}{\alpha }+2d=3\text{ }...(2) \\
\end{align}\]
Then, from (1) and (2), we get
\[\begin{align}
  & \dfrac{1}{\alpha }+d+d=3 \\
 & \Rightarrow 2+d=3 \\
 & \Rightarrow d=3-2=1 \\
\end{align}\]
Substituting obtained value in (1), we get
\[\begin{align}
  & \dfrac{1}{\alpha }+d=2 \\
 & \Rightarrow \dfrac{1}{\alpha }+1=2 \\
 & \Rightarrow \dfrac{1}{\alpha }=1 \\
\end{align}\]
Thus, the other terms are
\[\begin{align}
  & \dfrac{1}{\beta }=\dfrac{1}{\alpha }+d=1+1=2 \\
 & \dfrac{1}{\gamma }=\dfrac{1}{\alpha }+2d=1+2(1)=3 \\
 & \dfrac{1}{\delta }=\dfrac{1}{\alpha }+3d=1+3(1)=4 \\
\end{align}\]
So, applying these values into the products of the roots of the given equations, we get
\[\begin{align}
  & A=\dfrac{1}{\alpha \gamma }=\dfrac{1}{\alpha }\times \dfrac{1}{\gamma }=1\times 3=3 \\
 & B=\dfrac{1}{\beta \delta }=\dfrac{1}{\beta }\times \dfrac{1}{\delta }=2\times 4=8 \\
 & \therefore A=3,B=8 \\
\end{align}\]

Option ‘A’ is correct

Note: Here we need to remember the rule, if the given terms are in H.P then their reciprocals are in A.P and vice versa. So, by using this concept, we can evaluate the required variables in the above question.