1950 AHSME Problems/Problem 3: Difference between revisions
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==Solution== | ==Solution== | ||
We can divide by 4 to get: | We can divide by 4 to get: | ||
< | <imath> x^{2}-2x+\dfrac{5}{4}=0.</imath> | ||
Using Vieta's | The Vieta's formula states that in quadratic equation <imath>ax^2+bx+c</imath>, the sum of the roots of the equation is <imath>-\frac{b}{a}</imath>. | ||
Using Vieta's formula, we find that the roots add to <imath>2</imath> or <imath>\boxed{\textbf{(E)}\ \text{None of these}}</imath>. | |||
==Solution 2== | |||
We can divide by 4 to get: | |||
<imath> x^{2}-2x+\dfrac{5}{4}=0.</imath> | |||
By using quadratic formula you can find the roots are \(x=\frac{8\pm \sqrt{-16}}{8}\) adding the roots together to get 2. | |||
or <imath>\boxed{\textbf{(E)}\ \text{None of these}}</imath>. | |||
~samma | |||
==See Also== | ==See Also== | ||
Latest revision as of 17:42, 9 November 2025
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 
.
Solution 2
We can divide by 4 to get:
By using quadratic formula you can find the roots are \(x=\frac{8\pm \sqrt{-16}}{8}\) adding the roots together to get 2.
or .
~samma
See Also
| 1950 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 2 |
Followed by Problem 4 | |
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