# 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 such that the two polynomials and have at least one root in common.

## Solution

Let this root be . Then we have

Now, if , then we're done, since this satisfies the problem's conditions. If , then we can divide both sides by to obtain . Substituting this value into the first polynomial gives

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

Therefore the only possible values of are and . Q.E.D.