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

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==Problem==
 
==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
 
  
==Solution==
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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 - 0. By Vieta's formulas, the sum of the roots is 2, therefore their average is 1 (A).  
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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>
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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>.
  
 
==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}}
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{{MAA Notice}}

Latest revision as of 12:35, 7 June 2021

Problem

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

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

Solution 1

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 $1\Rightarrow \boxed{A}$.

Solution 2

We know that for an equation $ax^2 + bx + c = 0$, the sum of the roots is $\frac{-b}{a}$. This means that the sum of the roots for $ax^2 - 2ax + b = 0$ is $\frac{2a}{a}=2$. The average is the sum of the two roots divided by two, so the average is $\frac22 = 1 \Rightarrow \boxed{A}$.

See also

2008 AMC 10B (ProblemsAnswer KeyResources)
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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