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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?

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    $\begingroup$ As written, it's obvious. Should this be an if and only if statement? $\endgroup$ – Mose Wintner Aug 2 '15 at 6:15
  • $\begingroup$ @MoseWintner Yes, perhaps if and only if (iff) would have been appropriate. Maybe they missed the extra 'f'. $\endgroup$ – Hungry Blue Dev Aug 2 '15 at 6:18
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    $\begingroup$ Maybe add the condition $p,q\not =0$. This is implicitly assumed when you talk about quadratic equations and without this the only-if part is not longer true since if $p=0$ we have that $x = -\frac{r}{q}$ is a common root for all $r,q$. $\endgroup$ – Winther Aug 2 '15 at 7:20
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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 $$\left.\begin{align} p s^2 + q s + r &= 0 \quad\to\quad p s^3 + q s^2 + r s \phantom{+p\;} = 0 \\ q s^2 + r s + p &= 0 \quad\to\quad \phantom{p s^3 +\, } q s^2 + r s + p = 0 \end{align} \quad\right\}\quad\to\quad p(s^3-1) = 0 \quad\to\quad s^3 = 1$$

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!

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    $\begingroup$ "Yes, we have caught the rarest of all creatures - an elegant!"... Indeed, one of the most elegant approaches I've ever seen. $\endgroup$ – Hungry Blue Dev Aug 9 '15 at 7:08
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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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HINT:

If $a$ is a common root,

$$pa^2+qa+r=0\ \ \ \ (1)$$

$$qa^2+ra+p=0\ \ \ \ (2)$$

Solve $(1),(2)$ for $a^2,a$ and use $a^2=(a)^2$

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If $p + q + r = 0 \to 1 $ is a root, otherwise both become one equation and so they surely have the same roots.

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  • $\begingroup$ Yes, but how do I obtain that? That is, how to obtain $(p+q+r)$ as a factor of something equated to $0$? $\endgroup$ – Hungry Blue Dev Aug 2 '15 at 6:17
  • $\begingroup$ No, in that case both equations have a common root, namely $1$. $\endgroup$ – DeepSea Aug 2 '15 at 6:19
  • $\begingroup$ Well, I get what you mean, but I'm not allowed to back calculate (at least not on paper, I can do it in rough). How do I do it in the tedious way? $\endgroup$ – Hungry Blue Dev Aug 2 '15 at 6:25
  • $\begingroup$ Substitute $1$ into both equations and get $p\cdot1^2+q\cdot 1 + r = 0=q\cdot 1^2+r\cdot1 + p \to 1$ is a root of both. $\endgroup$ – DeepSea Aug 2 '15 at 6:30
  • $\begingroup$ No no, I understand what you're saying about $1$. But I'm asking about a general root (like $\alpha$) before finding it's value. Assume that the last two conditions are not given. $\endgroup$ – Hungry Blue Dev Aug 2 '15 at 6:34

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