Difference between revisions of "1971 Canadian MO Problems/Problem 4"
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== Problem == | == Problem == | ||
− | Determine all real numbers <math>a</math> such that the two polynomials <math> | + | 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 == | + | == Solutions == |
+ | === Solution 1 === | ||
− | ---- | + | Let this root be <math>r</math>. Then we have |
− | + | ||
− | + | <center> | |
− | + | <math>\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>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> | ||
+ | <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>a </math> are <math>1 </math> and <math>-2 </math>. Q.E.D. | ||
+ | |||
+ | === Solution 2 === | ||
+ | |||
+ | 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>. | ||
+ | Dividing <math>su</math> by <math>st</math>, we have: | ||
+ | <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> | ||
+ | Substituting <math>u = at</math> into the above, we have: | ||
+ | |||
+ | <center> | ||
+ | <math> \begin{matrix} a+t &=& 1+ at\\ | ||
+ | at - a - t +1 &=& 0\\ | ||
+ | (a-1)(t-1) &=& 0 \end{matrix}</math> | ||
+ | </center> | ||
+ | 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. | ||
+ | |||
+ | Thus, the only possible a values are: <math>a = 1, -2</math>. | ||
+ | |||
+ | == See Also == | ||
+ | {{Old CanadaMO box|num-b=3|num-a=5|year=1971}} | ||
+ | [[Category:Intermediate Algebra Problems]] |
Latest revision as of 15:35, 20 January 2014
Problem
Determine all real numbers such that the two polynomials and have at least one root in common.
Solutions
Solution 1
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.
Solution 2
Let and where is the common root. From Vieta's Formulas, we have: and . We see that . Dividing by , we have: Also, we have: Substituting into the above, we have:
Thus either or . We check to see that is indeed a possible value to satisfy the requirements. If , then from , we have , and from , we have , which also satisfies the requirements.
Thus, the only possible a values are: .
See Also
1971 Canadian MO (Problems) | ||
Preceded by Problem 3 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • | Followed by Problem 5 |