1950 AHSME Problems/Problem 3: Difference between revisions

Revision as of 13:14, 19 February 2012

The sum of the roots of the equation $4x^{2}+5-8x=0$ is equal to:

$\textbf{(A)}\ 8\qquad\textbf{(B)}\ -5\qquad\textbf{(C)}\ -\frac{5}{4}\qquad\textbf{(D)}\ -2\qquad\textbf{(E)}\ \text{None of these}$

We can divide by 4 to get: $x^{2}-2x+\dfrac{5}{4}=0.$

Using Vieta's formulas, we find that the roots add to $2\ \text{or}\ \boxed{\textbf{(E)}\ \text{None of these}}$ .

1950 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 2	Followed by Problem 4
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

@@ Line 9: / Line 9: @@
 <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==
 {{AHSME box|year=1950|num-b=2|num-a=4}}