Question Statement:-
If $p,q,r$ be in H.P. and $p$ and $r$ are of same sign, prove that the roots of the equation $px^2+2qx+r=0$ will be imaginary.
My solution:-
As $p,q,r$ are in H.P. then the relation b/w the three is as follows: $$q=\dfrac{2pr}{p+r}$$
Now consider the discriminant of the equation $px^2+2qx+r=0$.
We get, $D=4q^2-4pr=4(q^2-pr)=4\left(\dfrac{4p^2r^2}{(p+r)^2}-pr\right)=-4pr\left(\dfrac{p-r}{p+r}\right)^2$
Now, as it is given that the sign of $p$ and $r$ is same, so $\boxed{pr\gt0}$ and as $p,q,r$ can be same too so we get $D\le0$
We can obviously see that the roots can be equal (and hence real too) or imaginary with the conditions given in the question.
Is it that the question misses to include the point that $p,q,r$ should be distinct.
For a quadratic equation:
$$\text{a}x^2+\text{b}x+\text{c}=0\space\Longleftrightarrow\space x=\frac{-\text{b}\pm\sqrt{\text{b}^2-4\text{a}\text{c}}}{2\text{a}}$$
So, when ($\text{r}=\text{c}$ and $\text{p}=\text{a}$):
$$\text{q}=\text{b}=\frac{4\text{a}\text{c}}{\text{a}+\text{c}}$$
We get, for the quadratic solution:
$$x=\frac{-\frac{4\text{a}\text{c}}{\text{a}+\text{c}}\pm\sqrt{\left(\frac{4\text{a}\text{c}}{\text{a}+\text{c}}\right)^2-4\text{a}\text{c}}}{2\text{a}}=-\frac{2\text{c}}{\text{a}+\text{c}}\pm\frac{\sqrt{-\frac{\text{a}\text{c}\left(\text{a}-\text{c}\right)^2}{\left(\text{a}+\text{c}\right)^2}}}{\text{a}}$$