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?
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:
$$q = k p, \quad r = k q, \quad p = k r \quad\to\quad p = k^3 p \quad\to\quad p(k^3-1) = 0 \quad\to\quad k^3 = 1$$


  Thus, since $k=1$, we have $p = q = r$.


If the quadratics don't (necessarily) have both roots in common, but do have root $s$ in common, then
$$\left.\begin{align}
p s^2 + q s + r &amp;= 0 \quad\to\quad p s^3 + q s^2 + r s \phantom{+p\;} = 0 \\
q s^2 + r s + p &amp;= 0 \quad\to\quad \phantom{p s^3 +\, } q s^2 + r s + p = 0
\end{align} \quad\right\}\quad\to\quad p(s^3-1) = 0 \quad\to\quad s^3 = 1$$


  Thus, since $s = 1$, substituting back into either polynomial gives $p+q+r=0$.


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


  $$q = p \omega^n \qquad r = p \omega^{-n}$$


If $s = \omega^n$, then substituting back into either quadratic, and multiplying-through by an appropriate power of $\omega^{n}$ for balance, gives


  $$p\omega^{n} + q + r \omega^{-n} = 0$$


Just-slightly-less-easy-but-nonetheless-peasy!
If $p + q + r = 0 \to 1 $ is a root, otherwise both become one equation and so they surely have the same roots.