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

(Solution)
(Solution)
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<math> x^{2}-2x+\dfrac{5}{4}=0.</math>
 
<math> x^{2}-2x+\dfrac{5}{4}=0.</math>
  
Using Vieta's formulas, we find that the roots add to <math>2 \text{or} \boxed{\textbf{(E)} \text{None of these}}</math>.
+
Using Vieta's formulas, we find that the roots add to <math>2\ \text{or}\ \boxed{\textbf{(E)} \text{None of these}}</math>.
  
 
==See Also==
 
==See Also==
  
 
{{AHSME box|year=1950|num-b=2|num-a=4}}
 
{{AHSME box|year=1950|num-b=2|num-a=4}}

Revision as of 13:14, 19 February 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 $2\ \text{or}\ \boxed{\textbf{(E)} \text{None of these}}$.

See Also

1950 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 2
Followed by
Problem 4
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30
All AHSME Problems and Solutions