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

(Created page with "== 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 <math> x^2+px +q =0</math> to be <math> ...")
 
(→‎Solution)
 
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== Problem ==
 
== Problem ==
  
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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  
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<math> \textbf{(A)}\ \frac{B^2 - 4AC}{A^2}\qquad  
−
\textbf{(B)}\ \frac{B^2 \minus{} 2AC}{A^2}\qquad  
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\textbf{(B)}\ \frac{B^2 - 2AC}{A^2}\qquad  
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\textbf{(C)}\ \frac{2AC \minus{} B^2}{A^2}\qquad \\
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\textbf{(C)}\ \frac{2AC - B^2}{A^2}\qquad \\
−
\textbf{(D)}\ B^2 \minus{} 2C\qquad  
+
\textbf{(D)}\ B^2 - 2C\qquad  
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\textbf{(E)}\ 2C \minus{} B^2</math>
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\textbf{(E)}\ 2C - B^2</math>
  
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== Solution ==
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By Vieta's, <math>r + s = -\frac{B}{A}</math>, <math>rs = \frac{C}{A}</math>, and <math>r^2 + s^2 = -p</math>. Note that <math>(r+s)^2 = r^2 + s^2 + 2rs</math>.
 +
 +
Therefore, <math>(r + s)^2 - 2rs = r^2 + s^2</math>, or <math>-\left(\frac{B}{A}\right)^2 - \frac{2C}{A} = -p</math>.
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 +
Simplifying, <math>\frac{B^2 - 2CA}{A^2} = -p</math>.
  
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== Solution ==
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Finally, multiply both sides by <math>-1</math> to get <math>p = \frac{2CA - B^2}{A^2}</math>, making the answer <math>\fbox{C}</math>.
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<math>\fbox{}</math>
 
  
 
== See Also ==
 
== See Also ==

Latest revision as of 14:10, 22 February 2018

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

By Vieta's, $r + s = -\frac{B}{A}$, $rs = \frac{C}{A}$, and $r^2 + s^2 = -p$. Note that $(r+s)^2 = r^2 + s^2 + 2rs$.

Therefore, $(r + s)^2 - 2rs = r^2 + s^2$, or $-\left(\frac{B}{A}\right)^2 - \frac{2C}{A} = -p$.

Simplifying, $\frac{B^2 - 2CA}{A^2} = -p$.

Finally, multiply both sides by $-1$ to get $p = \frac{2CA - B^2}{A^2}$, making the answer $\fbox{C}$.

See Also

1958 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 40 		Followed by
Problem 42
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All AHSME Problems and Solutions

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