Difference between revisions of "1971 Canadian MO Problems/Problem 4"

(Problem)
(Added solution and category tag)
Line 2: Line 2:
 
Determine all real numbers <math>\displaystyle a</math> such that the two polynomials <math>\displaystyle x^2+ax+1</math> and <math>\displaystyle x^2+x+a</math> have at least one root in common.
 
Determine all real numbers <math>\displaystyle a</math> such that the two polynomials <math>\displaystyle x^2+ax+1</math> and <math>\displaystyle x^2+x+a</math> have at least one root in common.
  
== Solution ==  
+
== Solution ==
 +
 
 +
Let this root be <math>\displaystyle r</math>.  Then we have
 +
 
 +
<center>
 +
<math>\displaystyle \begin{matrix} r^2 + ar + 1 &=& r^2 + r + a\\
 +
ar + 1 &=& r + a\\
 +
(a-1)r &=& (a-1)\end{matrix} </math>
 +
</center>
 +
 
 +
Now, if <math>\displaystyle a = 1 </math>, then we're done, since this satisfies the problem's conditions.  If <math>\displaystyle a \neq 1</math>, then we can divide both sides by <math>\displaystyle (a - 1) </math> to obtain <math>\displaystyle r = 1 </math>.  Substituting this value into the first polynomial gives
 +
 
 +
<center>
 +
<math> \begin{matrix} 1 + a + 1 &=& 0\\
 +
a &=& -2 \end{matrix} </math>
 +
</center>
 +
It is easy to see that this value works for the second polynomial as well.
 +
 
 +
Therefore the only possible values of <math>\displaystyle a </math> are <math>\displaystyle 1 </math> and <math>\displaystyle -2 </math>.  Q.E.D.
  
 
----
 
----
Line 8: Line 26:
 
* [[1971 Canadian MO Problems/Problem 5|Next Problem]]
 
* [[1971 Canadian MO Problems/Problem 5|Next Problem]]
 
* [[1971 Canadian MO Problems|Back to Exam]]
 
* [[1971 Canadian MO Problems|Back to Exam]]
 +
 +
[[Category:Intermediate Algebra Problems]]

Revision as of 23:49, 27 July 2006

Problem

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

Solution

Let this root be $\displaystyle r$. Then we have

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

Now, if $\displaystyle a = 1$, then we're done, since this satisfies the problem's conditions. If $\displaystyle a \neq 1$, then we can divide both sides by $\displaystyle (a - 1)$ to obtain $\displaystyle r = 1$. 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 $\displaystyle a$ are $\displaystyle 1$ and $\displaystyle -2$. Q.E.D.


Invalid username
Login to AoPS