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1969 AHSME Problems/Problem 7

Revision as of 17:14, 30 September 2014 by Timneh (talk | contribs) (Created page with "== Problem == If the points <math>(1,y_1)</math> and <math>(-1,y_2)</math> lie on the graph of <math>y=ax^2+bx+c</math>, and <math>y_1-y_2=-6</math>, then <math>b</math> equals:...")
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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

$\fbox{A}$

See also

{{AHSME box|year=1969|num-b=6|num-a=28} The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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