https://artofproblemsolving.com/wiki/index.php?title=1964_AHSME_Problems/Problem_7&feed=atom&action=history1964 AHSME Problems/Problem 7 - Revision history2024-03-28T18:33:37ZRevision history for this page on the wikiMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=1964_AHSME_Problems/Problem_7&diff=107818&oldid=prevTalkinaway: Created page with "== Problem == Let n be the number of real values of <math>p</math> for which the roots of <math>x^2-px+p=0</math> are equal. Then n equals: <math>\textbf{(A)}\ 0 \qquad \te..."2019-07-23T16:03:31Z<p>Created page with "== Problem == Let n be the number of real values of <math>p</math> for which the roots of <math>x^2-px+p=0</math> are equal. Then n equals: <math>\textbf{(A)}\ 0 \qquad \te..."</p>
<p><b>New page</b></p><div>== Problem ==<br />
<br />
Let n be the number of real values of <math>p</math> for which the roots of <br />
<math>x^2-px+p=0</math><br />
are equal. Then n equals:<br />
<br />
<math>\textbf{(A)}\ 0 \qquad<br />
\textbf{(B)}\ 1 \qquad<br />
\textbf{(C)}\ 2 \qquad<br />
\textbf{(D)}\ \text{a finite number greater than 2}\qquad<br />
\textbf{(E)}\ \infty </math> <br />
<br />
== Solution ==<br />
<br />
If the roots of the quadratic <math>Ax^2 + Bx + C = 0</math> are equal, then <math>B^2 - 4AC = 0</math>. Plugging in <math>A=1, B=-p, C = p</math> into the equation gives <math>p^2 - 4p = 0</math>. This leads to <math>p = 0, 4</math>, so there are two values of <math>p</math> that work, giving answer <math>\boxed{\textbf{(C)}}</math>.<br />
<br />
==See Also==<br />
{{AHSME 40p box|year=1964|num-b=6|num-a=8}}<br />
<br />
[[Category:Introductory Algebra Problems]]<br />
<br />
{{MAA Notice}}</div>Talkinaway