Difference between revisions of "1958 AHSME Problems/Problem 33"

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== Problem ==
 
== Problem ==
For one root of <math> ax^2 \plus{} bx \plus{} c \equal{} 0</math> to be double the other, the coefficients <math> a,\,b,\,c</math> must be related as follows:
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For one root of <math> ax^2 + bx + c = 0</math> to be double the other, the coefficients <math> a,\,b,\,c</math> must be related as follows:
  
<math> \textbf{(A)}\ 4b^2 \equal{} 9c\qquad  
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<math> \textbf{(A)}\ 4b^2 = 9c\qquad  
\textbf{(B)}\ 2b^2 \equal{} 9ac\qquad  
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\textbf{(B)}\ 2b^2 = 9ac\qquad  
\textbf{(C)}\ 2b^2 \equal{} 9a\qquad \\
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\textbf{(C)}\ 2b^2 = 9a\qquad \\
\textbf{(D)}\ b^2 \minus{} 8ac \equal{} 0\qquad  
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\textbf{(D)}\ b^2 - 8ac = 0\qquad  
\textbf{(E)}\ 9b^2 \equal{} 2ac</math>
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\textbf{(E)}\ 9b^2 = 2ac</math>
  
 
== Solution ==
 
== Solution ==

Latest revision as of 23:21, 13 March 2015

Problem

For one root of $ax^2 + bx + c = 0$ to be double the other, the coefficients $a,\,b,\,c$ must be related as follows:

$\textbf{(A)}\ 4b^2 = 9c\qquad  \textbf{(B)}\ 2b^2 = 9ac\qquad  \textbf{(C)}\ 2b^2 = 9a\qquad \\ \textbf{(D)}\ b^2 - 8ac = 0\qquad  \textbf{(E)}\ 9b^2 = 2ac$

Solution

$\fbox{}$

See Also

1958 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 32
Followed by
Problem 34
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