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 "1965 AHSME Problems/Problem 22"

m (formatting fix)
(→‎Solution)
 
Line 12: Line 12:
 
== Solution ==
 
== Solution ==
  
−
<math>\fbox{A}</math>
+
<math>a_0</math> cannot be <math>0</math>, because <math>a_2 \neq 0</math>, so the polynomial can take non-zero values when <math>a_0=0</math> and thereby not satisfy the equation. Expanding <math>a_0(1-\frac{x}{r})(1-\frac{x}{s})</math> gives us <math>a_0(1-\frac{x}{r}-\frac{x}{s}+\frac{x^2}{rs})</math>. Using [[Vieta's formulas]], we see that <math>rs=\frac{a_0}{a_2}</math>, so <math>a_0=a_2rs</math>. Plugging this into the expanded right hand side of the given equation, we see that that side equals <math>a_0-a_2(r+s)x+a_2x^2</math>. Using Vieta's formulas again, we equate this expression to <math>a_0+a_1x+a_2x^2</math>, which is the left hand side of the given equation. Thus, as long as <math>a_0 \neq 0</math>, the equation holds for all values of <math>x</math>. This fact corresponds with answer choice <math>\fbox{\textbf{(A)}}</math>.
  
 
== See Also ==
 
== See Also ==

Latest revision as of 17:50, 18 July 2024

Problem

If $a_2 \neq 0$ and $r$ and $s$ are the roots of $a_0 + a_1x + a_2x^2 = 0$, then the equality $a_0 + a_1x + a_2x^2 = a_0\left (1 - \frac {x}{r} \right ) \left (1 - \frac {x}{s} \right )$ holds:

$\textbf{(A)}\ \text{for all values of }x, a_0\neq 0 \qquad \textbf{(B) }\ \text{for all values of }x \\ \textbf{(C) }\ \text{only when }x = 0 \qquad \textbf{(D) }\ \text{only when }x = r \text{ or }x = s \\ \textbf{(E) }\ \text{only when }x = r \text{ or }x = s, a_0 \neq 0$

Solution

$a_0$ cannot be $0$, because $a_2 \neq 0$, so the polynomial can take non-zero values when $a_0=0$ and thereby not satisfy the equation. Expanding $a_0(1-\frac{x}{r})(1-\frac{x}{s})$ gives us $a_0(1-\frac{x}{r}-\frac{x}{s}+\frac{x^2}{rs})$. Using Vieta's formulas, we see that $rs=\frac{a_0}{a_2}$, so $a_0=a_2rs$. Plugging this into the expanded right hand side of the given equation, we see that that side equals $a_0-a_2(r+s)x+a_2x^2$. Using Vieta's formulas again, we equate this expression to $a_0+a_1x+a_2x^2$, which is the left hand side of the given equation. Thus, as long as $a_0 \neq 0$, the equation holds for all values of $x$. This fact corresponds with answer choice $\fbox{\textbf{(A)}}$.

See Also

1965 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 21 		Followed by
Problem 23
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40
All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1965_AHSME_Problems/Problem_22&oldid=223044"