Difference between revisions of "1975 AHSME Problems/Problem 13"
Hashtagmath (talk | contribs) (Created page with "==Problem== The equation <math>x^6 - 3x^5 - 6x^3 - x + 8 = 0</math> has <math> \textbf{(A)} \text{ no real roots} \\ \textbf{(B)} \text{ exactly two distinct negative roots}...") |
(Added solution) |
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\textbf{(B)} \text{ exactly two distinct negative roots} \\ | \textbf{(B)} \text{ exactly two distinct negative roots} \\ | ||
\textbf{(C)} \text{ exactly one negative root} \\ | \textbf{(C)} \text{ exactly one negative root} \\ | ||
| − | \textbf{(D)} \text{ no negative roots, but at least one positive | + | \textbf{(D)} \text{ no negative roots, but at least one positive root} \\ |
\textbf{(E)} \text{ none of these} | \textbf{(E)} \text{ none of these} | ||
</math> | </math> | ||
==Solution== | ==Solution== | ||
| − | + | Let <math>P(x) = x^6 - 3x^5 - 6x^3 - x + 8</math>. When <math>x < 0</math>, <math>P(x) > 0</math>. Therefore, there are no negative roots. | |
| + | |||
| + | Notice that <math>P(1) = -1</math> and <math>P(0) = 8</math>. There must be at least one positive root between 0 and 1, therefore the answer is <math>\boxed{\textbf{(D)}}</math>. | ||
==See Also== | ==See Also== | ||
{{AHSME box|year=1975|num-b=12|num-a=14}} | {{AHSME box|year=1975|num-b=12|num-a=14}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Latest revision as of 13:51, 5 January 2022
Problem
The equation 
has
Solution
Let 
. When 
, 
. Therefore, there are no negative roots.
Notice that 
and 
. There must be at least one positive root between 0 and 1, therefore the answer is 
.
See Also
| 1975 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 12 |
Followed by Problem 14 | |
| 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 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
