Difference between revisions of "1950 AHSME Problems/Problem 3"
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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 is equal to:
Solution
We can divide by 4 to get:
Using Vieta's formulas, we find that the roots add to .
See Also
1950 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 2 |
Followed by Problem 4 | |
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All AHSME Problems and Solutions |