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Let $P,Q,R$ be positive real numbers, not all equal, such that some two of the below equation have exactly one common root, alpha. Then prove that alpha is real and negative and one of the below mentioned equation has imaginary roots Equations:- $Px^2 + 2Qx +R = 0$ , $Qx^2 + 2Rx + P =0$ , $Rx^2 + 2Px + Q =0$

My work : I used the condition for one common root which is $(c_1a_2 - c_2a_1)^2 = (a_1b_2 - a_2b_1)(b_1c_2 - b_2c_1) $ to all the three equations and then found discriminants. But don't know how to use them to proceed.

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$px^2+2qx+r=0\tag1$
$qx^2+2rx+p=0\tag2$
$rx^2+2px+q=0\tag3$

Let (1) and (2) have the common root $\alpha$.

We can write $(x-\alpha)(x-\beta)$ for (1) and use Vieta's formulas to see that $\frac{-2q}p = \alpha+\beta$ and $\frac rp = \alpha\beta$.

Similarly we can write $(x-\alpha)(x-\gamma)$ for (2) and observe that $\frac{-2r}q = \alpha+\gamma$ and $\frac pq = \alpha\gamma$.

Eliminating $\alpha$ from these results in $$\gamma-\beta = \dfrac{2}{pq}(q^2-pr) \quad \text{ and } \quad\frac\gamma \beta = \dfrac{p^2}{qr},$$

and solving these gives $$\beta = \dfrac{2r(q^2-pr)}{p(p^2-qr)} \quad \text{ and }\quad \gamma = \dfrac{2p(q^2-pr)}{q(p^2-qr)}.$$

Substituting into $\frac rp = \alpha\beta$ above gives us $$\alpha = \dfrac{qr(p^2-qr)}{2p^2(q^2-pr)}.$$

Because $p, q, r$ are real, $\alpha$ is real as well (as are $\beta$ and $\gamma$).

Now that we know that $\alpha, \beta$ and $\gamma$ are real we can say something about the discriminants of the quadratics.

For (1), $\quad 4q^2-4pr \ge 0 $ or $\frac{q^2}r \ge p$.

For (2), $\quad 4r^2-4pq \ge 0 $ or $\frac{r^2}q \ge p$.

If we assume WLOG that $r > q$ these inequalities lead to $p < q < r$, i.e. $p$ is the smallest.

The discriminant of (3) is $4p^2-4pq$ which is negative by the above inequality. Thus the roots of (3) are complex. This discriminant also serves to show that $\alpha < 0$ because $p^2-qr < 0$.

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