Difference between revisions of "1965 AHSME Problems/Problem 9"
m (fixed typo) |
m (maa notice) |
||
| Line 24: | Line 24: | ||
==See Also== | ==See Also== | ||
{{AHSME 40p box|year=1965|num-b=8|num-a=10}} | {{AHSME 40p box|year=1965|num-b=8|num-a=10}} | ||
| + | {{MAA Notice}} | ||
[[Category:Introductory Algebra Problems]] | [[Category:Introductory Algebra Problems]] | ||
Latest revision as of 17:03, 18 July 2024
Problem 9
The vertex of the parabola 
will be a point on the 
-axis if the value of 
is:
Solution
Notice that if the vertex of a parabola is on the x-axis, then the x-coordinate of the vertex must be a solution to the quadratic. Since the quadratic is strictly increasing on either side of the vertex, the solution must have double multiplicity, or the quadratic is a perfect square trinomial. This means that for the vertex of to be on the x-axis, the trinomial must be a perfect square, and have discriminant of zero. So,
Therefore 
, and our answer is 
.
See Also
| 1965 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 8 |
Followed by Problem 10 | |
| 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. 
