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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>
  
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Using Vieta's formulas, we find that the roots add to <math>2</math> or <math>\boxed{\textbf{(E)}\ \text{None of these}}</math>.
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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>.
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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==
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[[Category:Introductory Algebra Problems]]
 
[[Category:Introductory Algebra Problems]]
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{{MAA Notice}}

Latest revision as of 01:05, 10 June 2020

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: $x^{2}-2x+\dfrac{5}{4}=0.$

The Vieta's formula states that in quadratic equation $ax^2+bx+c$, the sum of the roots of the equation is $-\frac{b}{a}$. Using Vieta's formula, we find that the roots add to $2$ or $\boxed{\textbf{(E)}\ \text{None of these}}$.

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

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