1971 Canadian MO Problems/Problem 4

Problem

Determine all real numbers $a$ such that the two polynomials $x^2+ax+1$ and $x^2+x+a$ have at least one root in common.

Solutions

Solution 1

Let this root be $r$ . Then we have

$\begin{matrix} r^2 + ar + 1 &=& r^2 + r + a\\ ar + 1 &=& r + a\\ (a-1)r &=& (a-1)\end{matrix}$

Now, if $a = 1$ , then we're done, since this satisfies the problem's conditions. If $a \neq 1$ , then we can divide both sides by $(a - 1)$ to obtain $r = 1$ . Substituting this value into the first polynomial gives

$\begin{matrix} 1 + a + 1 &=& 0\\ a &=& -2 \end{matrix}$

It is easy to see that this value works for the second polynomial as well.

Therefore the only possible values of $a$ are $1$ and $-2$ . Q.E.D.

Solution 2

Let $x^2+ax+1 = (x-s)(x-t)$ and $x^2+x+a = (x-s)(x-u)$ where $s$ is the common root. From Vieta's Formulas, we have: $-(s+t) = a, -(s+u) = 1, st = 1,$ and $su = 1$ . We see that $s,t,u \neq 0$ . Dividing $su$ by $st$ , we have: $\frac{su}{st} = \frac{a}{1} \Rightarrow u = at$ Also, we have: $a+t = -s = 1+u \Rightarrow a+t = 1+u$ Substituting $u = at$ into the above, we have:

$\begin{matrix} a+t &=& 1+ at\\ at - a - t +1 &=& 0\\ (a-1)(t-1) &=& 0 \end{matrix}$

Thus either $a = 1$ or $t = 1$ . We check to see that $a = 1$ is indeed a possible value to satisfy the requirements. If $t = 1$ , then from $st = 1$ , we have $s = t = 1$ , and from $-(s+t) = a$ , we have $a = -(1+1) = -2$ , which also satisfies the requirements.