The quadratic polynomial $px^2-qx+r=0,$ where $p<0$ has one root positive and the other root negative,then

Question

The quadratic polynomial $px^2-qx+r=0,$ where $p<0$ has one root positive and the other root negative and magnitude of positive root is greater than magnitude of negative root.then
$(A)r^2-4q<0$
$(B)r^2-4p<0$
$(C)p+q>0$
$(D)r-p-q>0$

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Let $f(x)=px^2-qx+r$
As $p<0$ and one root is positive and the other root is negative.This show that $f(0)=r>0$.
One more condition is given,magnitude of positive root is greater than magnitude of negative root. But how do i choose the correct option which is given to be $(D)$ in answers.

Answer 1

D is not right answer. For example $-x^2-10x+1=0$, where $p=-1,q=10,r=1$ has two roots one of them positive and another negative and $r-p-q=1+1-10<0$.

There are no right variants. for (A), (B) and (C): $-x^2+10x+1=0$ where $p=-1,q=-10,r=1$ then $r^2-4q=41>0$, $r^2-4p=5>0$, $p+q=-11<0$.

Answer 2

It can not be D).

For example take $$-x^2-3x+1=0.$$

Answer 3

I claim that $\color{green}{r>0}$ is a necessary and sufficient condition.

If $r>0$, $q^2-4pr$ is positive and there are two real roots; furthermore, $\dfrac rp<0$ so that by Vieta the product of these roots is negative.

Reciprocally, two real roots of opposite sign yield $\dfrac rp<0$, hence $r>0$.

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Notice that none of A), B), C) nor D) do imply $r>0$.

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Update (edited question):

The condition on the magnitudes implies that the sum of the roots is positive, which translates to $\dfrac qp>0$ so that $q<0$ and the condition D) certainly holds and the others not.

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The necessary and sufficient condition is

$$\color{green}{r>0\land q<0}.$$