If the equation $x^2+qx+rp=0$ and $x^2+rx+pq=0$ have a common root, the other root will satisfy which of the equations

Question

If the equation $x^2+qx+rp=0$ and $x^2+rx+pq=0$ have a common root, the other root will satisfy which of the following equations?

a) $x^2-x+r=0$
b) $x^2+rx+p=0$
c) $x^2-2x+pr=0$
d) $x^2+px+qr=0$

MY WORK: I've attached what I tried for this question in the image.

I've tried to use the common root condition to arrive at a few equations(conditions) but I couldn't discover by myself how to arrive at the answer. I tried to put the value of equation from common root condition in the 2 quadratic equations but that was too lengthy and I didnt end up with the answer.

Answer 1

Hint: By inspection, the common root here is $p$.

Then from theorem of viete on both quadratics, you have

$$p+r = -q$$

Answer 2

Let $a$ be the common root

$$a^2+aq+rp=0$$

$$a^2+ra+pq=0$$

On subtraction, $$a(q-r)=p(q-r)$$

If $q=r,$ the two equations become identical, so both roots will be same.

$\implies q\ne r, a=p$

So, if $a,b$ are the roots the first equation, $$ab=rp\implies b=?$$

Similarly, if $a,c$ are the roots the second equation, $c=?$