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

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==Solution 2==
 
==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>-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</math>.
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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==
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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}}
 
{{MAA Notice}}

Latest revision as of 15:59, 13 April 2024

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 $\frac22 = 1 \Rightarrow \boxed{A}$.

Solution 3

Using the quadratic formula(assuming $a=1$),we get $x = \frac{2 \pm \sqrt{4-4b}}{2}$. Let $y = \sqrt{4-4b}$. Simplifying, we get $1 \pm \frac{y}{2}$. That means the sum of the solutions is $(1+\frac{y}{2}) + (1-\frac{y}{2}) = 2$, so the average is $2/2 = \boxed{\textbf{(A) } 1}$.

~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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