Difference between revisions of "2008 AMC 10B Problems/Problem 9"

Revision as of 14:16, 11 February 2009

A quadratic equation $ax^2 - 2ax + b = 0$ has two real solutions. What is the average of these two solutions?

$\mathrm{(A)}\ 1\qquad\mathrm{(B)}\ 2\qquad\mathrm{(C)}\ \frac ba\qquad\mathrm{(D)}\ \frac{2b}a\qquad\mathrm{(E)}\ \sqrt{2b-a}$

Dividing both sides by $a$ , we get $x^2 - 2x + b/a = 0$ . By Vieta's formulas, the sum of the roots is $2$ , therefore their average is $1$ .

2008 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 8	Followed by Problem 10
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
All AMC 10 Problems and Solutions

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 ==Problem==
-A quadratic equation ax^2 - 2ax + b = 0 has two real solutions. What is the average of these two solutions?
-A) 1 B) 2 C) b/a D) 2b/a
+A quadratic equation <math>ax^2 - 2ax + b = 0</math> has two real solutions. What is the average of these two solutions?
+<math>\mathrm{(A)}\ 1\qquad\mathrm{(B)}\ 2\qquad\mathrm{(C)}\ \frac ba\qquad\mathrm{(D)}\ \frac{2b}a\qquad\mathrm{(E)}\ \sqrt{2b-a}</math>
 ==Solution==
-Dividing both sides by a, we get x^2 - 2x + b/a - 0. By Vieta's formulas, the sum of the roots is 2, therefore their average is 1 (A).
+Dividing both sides by <math>a</math>, we get <math>x^2 - 2x + b/a = 0</math>. By Vieta's formulas, the sum of the roots is <math>2</math>, therefore their average is <math>1</math>.
 ==See also==
 {{AMC10 box|year=2008|ab=B|num-b=8|num-a=10}}