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

Latest revision as of 22:20, 18 January 2023

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 $x^2+x+a = (x-s)(x-u)$ 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: $a = 1, -2$ .

@@ Line 2: / Line 2: @@
 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
@@ Line 22: / Line 23: @@
 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-u)</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]]

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

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