Difference between revisions of "1950 AHSME Problems/Problem 3"

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==See Also==
 
==See Also==
  
{{AHSME box|year=1950|num-b=2|num-a=4}}
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{{AHSME 50p box|year=1950|num-b=2|num-a=4}}
  
 
[[Category:Introductory Algebra Problems]]
 
[[Category:Introductory Algebra Problems]]

Revision as of 04:39, 23 April 2012

Problem

The sum of the roots of the equation $4x^{2}+5-8x=0$ is equal to:

$\textbf{(A)}\ 8\qquad\textbf{(B)}\ -5\qquad\textbf{(C)}\ -\frac{5}{4}\qquad\textbf{(D)}\ -2\qquad\textbf{(E)}\ \text{None of these}$

Solution

We can divide by 4 to get: $x^{2}-2x+\dfrac{5}{4}=0.$

Using Vieta's formulas, we find that the roots add to $2$ or $\boxed{\textbf{(E)}\ \text{None of these}}$.

See Also

1950 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 2
Followed by
Problem 4
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All AHSME Problems and Solutions