Difference between revisions of "1965 AHSME Problems/Problem 22"
(created solution page) |
m (formatting fix) |
||
| Line 4: | Line 4: | ||
<math>a_0 + a_1x + a_2x^2 = a_0\left (1 - \frac {x}{r} \right ) \left (1 - \frac {x}{s} \right )</math> holds: | <math>a_0 + a_1x + a_2x^2 = a_0\left (1 - \frac {x}{r} \right ) \left (1 - \frac {x}{s} \right )</math> holds: | ||
| − | <math>\textbf{(A)}\ \text{for all values of }x, a_0\neq 0 | + | <math>\textbf{(A)}\ \text{for all values of }x, a_0\neq 0 \qquad |
\textbf{(B) }\ \text{for all values of }x \\ | \textbf{(B) }\ \text{for all values of }x \\ | ||
| − | \textbf{(C) }\ \text{only when }x = 0 | + | \textbf{(C) }\ \text{only when }x = 0 \qquad |
\textbf{(D) }\ \text{only when }x = r \text{ or }x = s \\ | \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 </math> | + | \textbf{(E) }\ \text{only when }x = r \text{ or }x = s, a_0 \neq 0 </math> |
== Solution == | == Solution == | ||
Revision as of 16:41, 18 July 2024
Problem
If 
and 
and 
are the roots of 
, then the equality
holds:
Solution
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. 
