1965 AHSME Problems/Problem 22
Problem
If and and are the roots of , then the equality holds:
Solution
cannot be , because , so the polynomial can take non-zero values when and thereby not satisfy the equation. Expanding gives us . Using Vieta's formulas, we see that , so . Plugging this into the expanded right hand side of the given equation, we see that that side equals . Using Vieta's formulas again, we equate this expression to , which is the left hand side of the given equation. Thus, as long as , the equation holds for all values of . This fact corresponds with answer choice .
See Also
1965 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 21 |
Followed by Problem 23 | |
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