Difference between revisions of "2008 AMC 10B Problems/Problem 9"
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==Problem== | ==Problem== | ||
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− | ==Solution== | + | A quadratic equation <math>ax^2 - 2ax + b = 0</math> has two real solutions. What is the average of these two solutions? |
− | Dividing both sides by a, we get x^2 - 2x + b/a | + | |
+ | <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> | ||
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+ | ==Solution 1== | ||
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+ | 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>. | ||
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+ | ==Solution 2== | ||
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+ | 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>. | ||
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+ | ==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>. | ||
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+ | ~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}} |
Latest revision as of 15:59, 13 April 2024
Contents
[hide]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 |
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