Difference between revisions of "2008 AMC 10B Problems/Problem 9"
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<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> | <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== | + | ==Solution 1== |
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\Rightarrow \boxed{A}</math>. | 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\Rightarrow \boxed{A}</math>. | ||
| − | == | + | ==Solution 2== |
| − | We know that for an equation <math>ax^2 + bx + c = 0</math>, the sum of the roots is <math>-b | + | We know that for an equation <math>ax^2 + bx + c = 0</math>, the sum of the roots is <math>\frac{-b}{a}</math>. This means that the sum of the roots for <math>ax^2 - 2ax + b = 0</math> is <math>\frac{2a}{a}=2</math>. The average is the sum of the two roots divided by two, so the average is <math>\frac22 = 1 \Rightarrow \boxed{A}</math>. |
| + | |||
| + | ==Solution 3== | ||
| + | Using the quadratic formula(assuming <math>a=1</math>),we get <math>x = \frac{2 \pm \sqrt{4-4b}}{2}</math>. Let <math>y = \sqrt{4-4b}</math>. Simplifying, we get <math>1 \pm \frac{y}{2} </math>. That means the sum of the solutions is <math>(1+\frac{y}{2}) + (1-\frac{y}{2}) = 2</math>, so the average is <math>2/2 = \boxed{\textbf{(A) } 1}</math>. | ||
| + | |||
| + | ~idk12345678 | ||
==See also== | ==See also== | ||
{{AMC10 box|year=2008|ab=B|num-b=8|num-a=10}} | {{AMC10 box|year=2008|ab=B|num-b=8|num-a=10}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Latest revision as of 15:59, 13 April 2024
Problem
A quadratic equation 
has two real solutions. What is the average of these two solutions?
Solution 1
Dividing both sides by 
, we get 
. By Vieta's formulas, the sum of the roots is 
, therefore their average is 
.
Solution 2
We know that for an equation 
, the sum of the roots is 
. This means that the sum of the roots for is 
. The average is the sum of the two roots divided by two, so the average is 
.
Solution 3
Using the quadratic formula(assuming 
),we get 
. Let 
. Simplifying, we get 
. That means the sum of the solutions is 
, so the average is 
.
~idk12345678
See also
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
