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If $a, b$ and $c$ are in arithmetic progression, then the roots of the equation $a x^{2}-2 h x+c=0$ are
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Verified Answer
The correct answer is:
1 and $\frac{c}{a}$
Since, $a, b$ and $c$ are in $\mathrm{AP}$.
$\therefore$
$2 b=a+c$
Given, quadratic equation, $a x^{2}-2 b x+c=0$
$\Rightarrow \quad a x^{2}-(a+c) x+c=0$
$2 b=a+c)$
$\Rightarrow \quad a x^{2}-a x-c x+c=0$
$\begin{array}{lr}\Rightarrow & a x(x-1)-c(x-1)=0 \\ \Rightarrow & (x-1)(a x-c)=0 \\ \Rightarrow & x=1, \frac{c}{a}\end{array}$
$\therefore$
$2 b=a+c$
Given, quadratic equation, $a x^{2}-2 b x+c=0$
$\Rightarrow \quad a x^{2}-(a+c) x+c=0$
$2 b=a+c)$
$\Rightarrow \quad a x^{2}-a x-c x+c=0$
$\begin{array}{lr}\Rightarrow & a x(x-1)-c(x-1)=0 \\ \Rightarrow & (x-1)(a x-c)=0 \\ \Rightarrow & x=1, \frac{c}{a}\end{array}$
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