Difference between revisions of "1965 AHSME Problems/Problem 9"
(Created page with "== Problem 9== The vertex of the parabola <math>y = x^2 - 8x + c</math> will be a point on the <math>x</math>-axis if the value of <math>c</math> is: <math>\textbf{(A)}\ -...") |
m (maa notice) |
||
| (2 intermediate revisions by the same user not shown) | |||
| Line 20: | Line 20: | ||
\end{align*}</cmath> | \end{align*}</cmath> | ||
| − | Therefore <math>c=16</math>, and our answer is <math>\boxed{\textbf{(E)}}</math> | + | Therefore <math>c=16</math>, and our answer is <math>\boxed{\textbf{(E)}}</math>. |
| + | |||
| + | ==See Also== | ||
| + | {{AHSME 40p box|year=1965|num-b=8|num-a=10}} | ||
| + | {{MAA Notice}} | ||
| + | |||
| + | [[Category:Introductory Algebra Problems]] | ||
Latest revision as of 16:03, 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. 
