If $p, q, r$ are in $G.P.$ and the equations, $p x^{2}+2 q x+r=0$ and $d x^{2}+2 e x+f=0$ have a common root, then show that $\frac{d}{p}, \frac{e}{q}, \frac{f}{r}$ are in $A.P.$
If $p, q, r$ are in $G.P.$ and the equations, $p x^{2}+2 q x+r=0$ and $d x^{2}+2 e x+f=0$ have a common root, then show that $\frac{d}{p}, \frac{e}{q}, \frac{f}{r}$ are in $A.P.$
The equation $p x^{2}+2 q x+r=0$ has roots given by
$x=\frac{-2 q \pm \sqrt{4 q^{2}-4 r p}}{2 p}$
since $p, q, r$ are in $G.P.$ $q^{2}=p r .$ Thus $x=\frac{-q}{p}$ but $\frac{-q}{p}$ is also root of ${d{x^2} + 2ex + f = 0}$ (Why ?). Therefore
$d\left(\frac{-q}{p}\right)^{2}+2 e\left(\frac{-q}{p}\right)+f=0$
or $d q^{2}-2 e q p+f p^{2}=0$ .........$(1)$
Dividing $(1)$ by $pq^{2}$ and using $q^{2}=$ $pr,$ we get
$\frac{d}{p}-\frac{2 e}{q}+\frac{f p}{p r}=0,$ or $\quad \frac{2 e}{q}=\frac{d}{p}+\frac{f}{r}$
Hence $\frac{d}{p}, \frac{e}{q}, \frac{f}{r}$ are in $A.P.$
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