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

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== 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.
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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 ==
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== Solutions ==
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=== Solution 1 ===
  
Let this root be <math>\displaystyle r</math>.  Then we have
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Let this root be <math>r</math>.  Then we have
  
 
<center>
 
<center>
<math>\displaystyle \begin{matrix} r^2 + ar + 1 &=& r^2 + r + a\\
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<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
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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>
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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.
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Therefore the only possible values of <math>a </math> are <math>1 </math> and <math>-2 </math>.  Q.E.D.
  
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=== Solution 2 ===
* [[1971 Canadian MO Problems/Problem 3|Previous Problem]]
 
* [[1971 Canadian MO Problems/Problem 5|Next Problem]]
 
* [[1971 Canadian MO Problems|Back to Exam]]
 
  
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Let <math>x^2+ax+1 = (x-s)(x-t) </math> and <math> x^2+x+a = (x-s)(x-t)</math> where <math>s</math> is the common root. From Vieta's Formulas, we have: <math>-(s+t) = a,        -(s+u) = 1,      st = 1, </math> and <math>        su = 1</math>. We see that <math>s,t,u \neq 0</math>.
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Dividing <math>su</math> by <math>st</math>, we have:
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<cmath> \frac{su}{st} = \frac{a}{1} \Rightarrow u = at</cmath> Also, we have: <cmath>a+t = -s = 1+u \Rightarrow a+t = 1+u</cmath>
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Substituting <math>u = at</math> into the above, we have:
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<center>
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<math> \begin{matrix} a+t &=& 1+ at\\
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at - a - t +1 &=& 0\\
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(a-1)(t-1) &=& 0 \end{matrix}</math>
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</center>
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Thus either <math>a = 1</math> or <math>t = 1</math>. We check to see that <math>a = 1</math> is indeed a possible value to satisfy the requirements. If <math>t = 1</math>, then from <math>st = 1</math>, we have <math>s = t = 1</math>, and from <math>-(s+t) = a</math>, we have <math>a = -(1+1) = -2</math>, which also satisfies the requirements.
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Thus, the only possible a values are: <math>a = 1, -2</math>.
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== See Also ==
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{{Old CanadaMO box|num-b=3|num-a=5|year=1971}}
 
[[Category:Intermediate Algebra Problems]]
 
[[Category:Intermediate Algebra Problems]]

Latest revision as of 15:35, 20 January 2014

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 $x^2+x+a = (x-s)(x-t)$ 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.

Thus, the only possible a values are: $a = 1, -2$.

See Also

1971 Canadian MO (Problems)
Preceded by
Problem 3
1 2 3 4 5 6 7 8 Followed by
Problem 5
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