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 | + | The Vieta's formula states that in quadratic equation <math>ax^2+bx+c</math>, the sum of the roots of the equation is <math>-\frac{b}{a}</math>. |
+ | Using Vieta's formula, we find that the roots add to <math>2</math> or <math>\boxed{\textbf{(E)}\ \text{None of these}}</math>. | ||
==See Also== | ==See Also== | ||
− | {{AHSME box|year=1950|num-b=2|num-a=4}} | + | {{AHSME 50p box|year=1950|num-b=2|num-a=4}} |
+ | |||
+ | [[Category:Introductory Algebra Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 01:05, 10 June 2020
Problem
The sum of the roots of the equation is equal to:
Solution
We can divide by 4 to get:
The Vieta's formula states that in quadratic equation , the sum of the roots of the equation is . Using Vieta's formula, we find that the roots add to or .
See Also
1950 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 2 |
Followed by Problem 4 | |
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All AHSME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.