<s>[INST] Condition for common roots of two Quadratic equations: $px^2+qx+r=0$ and $qx^2+rx+p=0$ The question is:

Show that the equation $px^2+qx+r=0$ and $qx^2+rx+p=0$ will have a common root if $p+q+r=0$ or $p=q=r$.

How should I approach the problem? Should I assume three roots $\alpha$, $\beta$ and $\gamma$ (where $\alpha$ is the common root)? Or should I try combining these two and try to get a value for the Discriminant? Or should I do something else altogether?
 [/INST] If the two quadratic polynomials $f(x) = px^2 + qx + r$ and $g(x) = qx^2 + rx + p$ have a common root then this is also a root of the linear polynomial
$$h(x) = qf(x) - pg(x) = (q^2-pr)x - (p^2-qr)$$
which means that either $q^2-pr=0$ and $p^2-qr=0$ for which $p=q=r$ or that $x = \frac{p^2-qr}{q^2-pr}$ is the common root. Inserting this into  either $f(x)$ of $g(x)$ gives us
$$p^3 + q^3 + r^3 = 3pqr$$
Now
$$p^3+q^3 + r^2 - 3pqr = (p+q+r)\left[\frac{3(p^2+q^2+r^2) - (p+q+r)^2}{2}\right]$$
so $p+q+r = 0$ or $(p+q+r)^2 = 3(p^2+q^2+r^2)$. 
By Cauchy-Schwarz we have $(p+q+r)^2 \leq 3(p^2+q^2+r^2)$ with equality only when $p=q=r$ so the two polynomials have a common root if and only if $p+q+r=0$ or $p=q=r$.


${\bf Assumptions}$: This answer assumes $p,q\not=0$ and real polynomials; $p,q,r\in\mathbb{R}$. The first condition is assumed since the polynomials is said to be quadratic and otherwise the only-if statement is not true since $p=0\implies x = -\frac{r}{q}$ is a common root for all $q,r$ and if $q=0$ we need $p=r=0$ to have a common root. If the second condition is relaxed allowing for complex coefficients then there are other solutions as shown in the answer by Blue.

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