Question

If $p,q,r$ are in $G.P.$ and the equations, $p{x}^2+2qx+r=0$ and $d{x}^2+8ex+f=0$ have a common root, then show that $\frac{d}{p},\frac{e}{q},\frac{f}{r}$ are in $A.P.$

Answer 1

It is given that,

$p,q,rinG.P.$

So, thier common ratio is same

$\frac{q}{p}=\frac{r}{q}$

$q^2=rp---\left(1\right)$

solving the equation

$p{x}^2+2qx+r=0$

compare that

$a{x}^2+bx+c=0$

roots are

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

here, $a=\varphi ,b=2qandc=r$

$x=\frac{-2q\pm \sqrt{4q^2-4pr}}{2p}$

$x=\frac{-2q\pm \sqrt{4pr-4pr}}{2p}$

$x=\frac{-2q\pm 0}{2p}$

$x=-\frac{q}{p}$

Thus,

$x=-\frac{q}{p}$ is the root of the equation

$p{x}^2+2qx+r=0$

Also, given that equations $p{x}^2+2qx+r=0$

and $d{x}^2+2ex+f=0$ have a common root

So, $-\frac{q}{p}$ is a root of $d{x}^2+2ex+f=0$

putting $x=-\frac{q}{p}$ in $d{x}^2+2ex-f=0$

$d{\left(\frac{-q}{p}\right)}^2+2e\left(\frac{-q}{p}\right)+f=0$

$\frac{d{q}^2}{p^2}-\frac{2eq}{p}+f=0$

$d{q}^2-2eqp+f{p}^2=0---\left(2\right)$

But We need to show that

$\frac{d}{p},\frac{e}{q},\frac{f}{r}$ are in an A.P.

Now, from equation(2) and weget.

$d{q}^2-2eqp+f{p}^2=0$

On dividing this $p{q}^2$

$\frac{d{q}^2}{p{q}^2}-\frac{2epq}{p{q}^2}+\frac{f{p}^2}{p{q}^2}=\frac{0}{p{q}^2}$

$\frac{d}{p}-\frac{2e}{q}+\frac{fp}{q^2}=0$

$\frac{d}{p}+\frac{fp}{q^2}=\frac{2e}{q}$

putting $q^2=pr$

$\frac{d}{p}+\frac{fp}{pr}=\frac{2e}{q}$

$\frac{d}{p}+\frac{f}{r}=\frac{2e}{q}$

$\therefore \frac{d}{p},\frac{e}{q},\frac{f}{r},$ are in an A.P.

Hence proved.