Question

If $a,b$ and $c$ 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

The correct option is A

$1$ and $\frac{c}{a}$

Explanation for the correct option:

Step 1. Find the roots of the equation $a{x}^2-2bx+c=0$:

Let $ɑ,\beta$ be the roots of the equation $a{x}^2–2bx+c=0$.

$\Rightarrow$$ɑ+\beta =\frac{2b}{a},ɑ\beta =\frac{c}{a}$

It is given that $a,b$ and $c$ are in arithmetic progression, then

$2b=a+c$

Divide both sides by $a$

$\frac{2b}{a}=1+\frac{c}{a}$

$\Rightarrow$ $ɑ+\beta =1+ɑ\beta$

$\Rightarrow$$(\alpha -1)(\beta -1)=0$

$\Rightarrow$$ɑ=1$ or $\beta =1$

Step 2: If $ɑ=1$, then the product of the roots is:

$ɑ\beta =\frac{c}{a}$

$\Rightarrow$ $\beta =\frac{c}{a}$

$\therefore$The roots of the given equations are $\left(1,\frac{c}{a}\right)$

Hence, option ‘A’ is correct.