Difference between revisions of "1969 AHSME Problems/Problem 7"

(Solution to Problem 7)
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The answer is <math>\boxed{\textbf{(A)}}</math>.
 
The answer is <math>\boxed{\textbf{(A)}}</math>.
  
== See also ==
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== See Also ==
 
{{AHSME 35p box|year=1969|num-b=6|num-a=8}}
 
{{AHSME 35p box|year=1969|num-b=6|num-a=8}}
  
 
[[Category: Introductory Algebra Problems]]
 
[[Category: Introductory Algebra Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 16:01, 20 June 2018

Problem

If the points $(1,y_1)$ and $(-1,y_2)$ lie on the graph of $y=ax^2+bx+c$, and $y_1-y_2=-6$, then $b$ equals:

$\text{(A) } -3\quad \text{(B) } 0\quad \text{(C) } 3\quad \text{(D) } \sqrt{ac}\quad \text{(E) } \frac{a+c}{2}$

Solution

Because the two points are on the quadratic, $y_1 = a + b + c$ and $y_2 = a - b + c$. Because $y_1 - y_2 = -6$, \[(a+b+c)-(a-b+c)=-6\] \[2b=-6\] \[b=-3\] The answer is $\boxed{\textbf{(A)}}$.

See Also

1969 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 6
Followed by
Problem 8
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All AHSME Problems and Solutions

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