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> | ||
| − | 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}. | + | 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 < | + | 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== | ||
Latest revision as of 02:05, 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. 
