Question

If $a, b, c$ are in arithmetic progression, then the roots of the equation $ax^{2}-2bx+c=0$ are

• A. $1$ and $\frac{c}{a}$
• B. $-\frac{1}{a}$ and $-c$
• C. $-1$ and $-\frac{c}{a}$
• D. $-2$ and $-\frac{c}{2a}$

Answer 1

Answer: (A)

Since, $a, b$ and $c$ are in $AP$.
$\therefore 2 b=a +c$
Given, quadratic equation,
$a x^{2}-2 b x +c=0$
$\Rightarrow a x^{2}-(a +c) x +c=0\,\, (2 b=a +c)$
$\Rightarrow a x^{2}-a x -c x +c=0$
$\Rightarrow a x(x-1)-c(x-1) =0$
$\Rightarrow (x-1)(a x- c) =0$
$\Rightarrow x=1,\, \frac{c}{a}$