Question

If $a,b,c$ are in arithmetic progression, then the roots of the equation $a{x}^2-2bx+c=0$ 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 $AP$.
$\therefore 2b=a+c$
Given, quadratic equation,
$a{x}^2-2bx+c=0$
$\Rightarrow a{x}^2-(a+c)x+c=0(2b=a+c)$
$\Rightarrow a{x}^2-ax-cx+c=0$
$\Rightarrow ax(x-1)-c(x-1)=0$
$\Rightarrow (x-1)(ax-c)=0$
$\Rightarrow x=1,\frac{c}{a}$