1950 AHSME Problems/Problem 15
Contents
Problem
The real roots of are:
Solution
Solution 1
This looks similar to a difference of squares, so we can write it as Neither of these factors are real.
Also, looking at the answer choices, there is no way multiplying two polynomials of degree will result in a polynomial of degree as well. Therefore the real factors are
Solution 2
Let's try to find all real roots of . If there was a real root (call it ), then it would satisfy . Rearranging gives that . Therefore a real root of would satisfy . However, the Trivial Inequality states that no real satisfies , which must mean that there are no real roots ; they are
Solution 3 (Easier Solution)
Let us try to factor We can see that the first term has variables only and the second has a constant only. Therefore, we cannot factor out anything and our answer is
See Also
1950 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Problem 16 | |
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