Difference between revisions of "1958 AHSME Problems/Problem 41"

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== Problem ==
 
== Problem ==
  
The roots of <math> Ax^2 \plus{} Bx \plus{} C \equal{} 0</math> are <math> r</math> and <math> s</math>. For the roots of
+
The roots of <math> Ax^2 + Bx + C = 0</math> are <math> r</math> and <math> s</math>. For the roots of
 
<math> x^2+px +q =0</math>
 
<math> x^2+px +q =0</math>
  
 
to be <math> r^2</math> and <math> s^2</math>, <math> p</math> must equal:
 
to be <math> r^2</math> and <math> s^2</math>, <math> p</math> must equal:
  
<math> \textbf{(A)}\ \frac{B^2 \minus{} 4AC}{A^2}\qquad  
+
<math> \textbf{(A)}\ \frac{B^2 - 4AC}{A^2}\qquad  
\textbf{(B)}\ \frac{B^2 \minus{} 2AC}{A^2}\qquad  
+
\textbf{(B)}\ \frac{B^2 - 2AC}{A^2}\qquad  
\textbf{(C)}\ \frac{2AC \minus{} B^2}{A^2}\qquad \\
+
\textbf{(C)}\ \frac{2AC - B^2}{A^2}\qquad \\
\textbf{(D)}\ B^2 \minus{} 2C\qquad  
+
\textbf{(D)}\ B^2 - 2C\qquad  
\textbf{(E)}\ 2C \minus{} B^2</math>
+
\textbf{(E)}\ 2C - B^2</math>
 
 
  
 
== Solution ==
 
== Solution ==

Revision as of 23:25, 13 March 2015

Problem

The roots of $Ax^2 + Bx + C = 0$ are $r$ and $s$. For the roots of $x^2+px +q =0$

to be $r^2$ and $s^2$, $p$ must equal:

$\textbf{(A)}\ \frac{B^2 - 4AC}{A^2}\qquad  \textbf{(B)}\ \frac{B^2 - 2AC}{A^2}\qquad  \textbf{(C)}\ \frac{2AC - B^2}{A^2}\qquad \\ \textbf{(D)}\ B^2 - 2C\qquad  \textbf{(E)}\ 2C - B^2$

Solution

$\fbox{}$

See Also

1958 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 40
Followed by
Problem 42
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All AHSME Problems and Solutions

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