Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(Added solution and category tag)
(box)
Line 1: Line 1:
 
== Problem ==
 
== Problem ==
−
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>a</math> such that the two polynomials <math>x^2+ax+1</math> and <math>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
+
Let this root be <math>r</math>.  Then we have
  
 
<center>
 
<center>
−
<math>\displaystyle \begin{matrix} r^2 + ar + 1 &=& r^2 + r + a\\
+
<math>\begin{matrix} r^2 + ar + 1 &=& r^2 + r + a\\
 
ar + 1 &=& r + a\\
 
ar + 1 &=& r + a\\
 
(a-1)r &=& (a-1)\end{matrix} </math>
 
(a-1)r &=& (a-1)\end{matrix} </math>
 
</center>
 
</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
+
Now, if <math>a = 1 </math>, then we're done, since this satisfies the problem's conditions.  If <math>a \neq 1</math>, then we can divide both sides by <math>(a - 1) </math> to obtain <math>r = 1 </math>.  Substituting this value into the first polynomial gives
  
 
<center>
 
<center>
Line 20: Line 20:
 
It is easy to see that this value works for the second polynomial as well.
 
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.
+
Therefore the only possible values of <math>a </math> are <math>1 </math> and <math>-2 </math>.  Q.E.D.
−
 
 
−
----
 
−
* [[1971 Canadian MO Problems/Problem 3|Previous Problem]]
 
−
* [[1971 Canadian MO Problems/Problem 5|Next Problem]]
 
−
* [[1971 Canadian MO Problems|Back to Exam]]
 
  
 +
{{Old CanadaMO box|num-b=3|num-a=5|year=1971}}
 
[[Category:Intermediate Algebra Problems]]
 
[[Category:Intermediate Algebra Problems]]

Revision as of 21:48, 17 November 2007

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.

Solution

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.

1971 Canadian MO (Problems)
Preceded by
Problem 3 		1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • Followed by
Problem 5
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1971_Canadian_MO_Problems/Problem_4&oldid=19651"