Question

Given that $\alpha ,\gamma$ are the roots of the equation $A{x}^2-4x+1=0$ and $\beta ,\delta$ the roots of equation, $B{x}^2-6x+1=0$ where $\alpha ,\beta ,\gamma ,\delta ,$ are in H.P. Then

• A. A =3
• B. B = 8
• C. A = 8
• D. B = 3

Answer 1

Answer: (A), (B)

The correct options are
A A =3
B B = 8

$A{x}^2-4x+1=0\alpha +\gamma =\frac{4}{A},\alpha \gamma =\frac{1}{A}B{x}^2-6x+1=0\beta +\delta =\frac{6}{B},\beta \delta =\frac{1}{B}$
$\frac{1}{\alpha },\frac{1}{\beta },\frac{1}{\gamma },\frac{1}{\delta }$ are in A.P.
$\Rightarrow \frac{1}{\gamma }-\frac{1}{\alpha }=\frac{1}{\delta }-\frac{1}{\beta }\Rightarrow \frac{\frac{\sqrt{16-4A}}{|A|}}{\frac{1}{A}}=\frac{\frac{\sqrt{36-4B}}{|B|}}{\frac{1}{B}}\Rightarrow \sqrt{16-4A}=\sqrt{36-4B}(\because A,B>0)\Rightarrow A-B=-5\Rightarrow \frac{2}{\beta }=\frac{1}{\alpha }+\frac{1}{\gamma }=\frac{\frac{4}{A}}{\frac{1}{A}}=4\Rightarrow \beta =\frac{1}{2}$
$\Rightarrow \frac{B}{4}-3+1=0\Rightarrow B=8\Rightarrow A=3$