Question

If ${\beta }_1,{\beta }_3$ are roots of equation $a{x}^2-6x+1=0$ & ${\beta }_2,{\beta }_4$ are roots of the equation $c{x}^2-10x+1=0$. If ${\beta }_1,{\beta }_2,{\beta }_3,{\beta }_4$ are in H.P., then values of $a$ &$c$ respectively are

• A. $5,-21$
• B. $-5,-21$
• C. $5,21$
• D. $-5,21$

Answer 1

Answer: (C)

$5,21$

${\beta }_1,{\beta }_{3,}$ are roots of $a{x}^2-6x+1=0$ & ${\beta }_2,{\beta }_{4,}$ are roots of $a{x}^2-10x+1=0$
$\therefore {\beta }_1+{\beta }_3=\frac{6}{a}$ & $\therefore {\beta }_1{\beta }_3=\frac{1}{a}$
$\therefore \frac{{\beta }_1{\beta }_3}{{\beta }_1+{\beta }_3}=\frac{1}{3}$
and ${\beta }_2+{\beta }_4=\frac{10}{c}and{\beta }_2{\beta }_4=\frac{1}{c}$
$\frac{2{\beta }_2{\beta }_4}{{\beta }_2+{\beta }_4}=\frac{1}{5}$
Now ${\beta }_1,{\beta }_2,{\beta }_3,{\beta }_4\in HP$
$\therefore \frac{2{\beta }_1{\beta }_3}{{\beta }_1+{\beta }_3}=\frac{1}{3}and{\beta }_3=\frac{2{\beta }_2{\beta }_4}{{\beta }_2+{\beta }_4}=\frac{1}{5}$
$\because {\beta }_3$ is root of $a{x}^2-6x+1=0$
$\therefore a{\left(\frac{1}{5}\right)}^2-6\left(\frac{1}{5}\right)+1=0$
$\therefore \frac{a}{25}-\frac{1}{5}=0\therefore a=5$
and ${\beta }_2$ is root of $a{x}^2-10x+1=0where{\beta }_2=\frac{1}{3}$
$\therefore \frac{c}{9}-\frac{10}{3}+1=0$
$\therefore c=21$ $\therefore a=5,c=21$
Hence choice (c) is correct answer