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?
Best Answer
The "if" part is clear, so we'll deal with the "only if". Moreover, we take the polynomials to be "true" quadratics ---that is, $p$ and $q$ are non-zero--- since otherwise the proposition is false (as @Winther mentions in a comment to the OP).
First, note that quadratics with a common root could have both roots in common, in which case they are equivalent. This means that multiplying-through by some $k$ turns one quadratic equation into the other, coefficient-wise: $$q = k p, \quad r = k q, \quad p = k r \quad\to\quad p = k^3 p \quad\to\quad p(k^3-1) = 0 \quad\to\quad k^3 = 1$$
If the quadratics don't (necessarily) have both roots in common, but do have root $s$ in common, then $$\left.\begin{align} p s^2 + q s + r &= 0 \quad\to\quad p s^3 + q s^2 + r s \phantom{+p\;} = 0 \\ q s^2 + r s + p &= 0 \quad\to\quad \phantom{p s^3 +\, } q s^2 + r s + p = 0 \end{align} \quad\right\}\quad\to\quad p(s^3-1) = 0 \quad\to\quad s^3 = 1$$
Easy-peasy!
But wait ... The equations $k^3 = 1$ and $s^3 = 1$ have fully three solutions: namely, $\omega^{0}$, $\omega^{+1}$, $\omega^{-1}$, where $\omega = \exp(2i\pi/3) = (-1+i\sqrt{3})/2$. Nobody said coefficients $p$, $q$, $r$, or common root $s$, were real, did they?
If $k = \omega^n$, then we have in general that
If $s = \omega^n$, then substituting back into either quadratic, and multiplying-through by an appropriate power of $\omega^{n}$ for balance, gives
Just-slightly-less-easy-but-nonetheless-peasy!