Difference between revisions of "1969 AHSME Problems/Problem 7"
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== Solution == | == Solution == | ||
| − | <math>\ | + | Because the two points are on the quadratic, <math>y_1 = a + b + c</math> and <math>y_2 = a - b + c</math>. Because <math>y_1 - y_2 = -6</math>, |
| + | <cmath>(a+b+c)-(a-b+c)=-6</cmath> | ||
| + | <cmath>2b=-6</cmath> | ||
| + | <cmath>b=-3</cmath> | ||
| + | The answer is <math>\boxed{\textbf{(A)}}</math>. | ||
| − | == See | + | == See Also == |
| − | {{AHSME box|year=1969|num-b=6|num-a= | + | {{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 17:01, 20 June 2018
Problem
If the points 
and 
lie on the graph of 
, and 
, then 
equals:
Solution
Because the two points are on the quadratic, 
and 
. Because 
,
![\[(a+b+c)-(a-b+c)=-6\]](http://latex.artofproblemsolving.com/1/a/7/1a7233b34481be5cdd3b1e9e8b4143c12ab2110b.png)
![\[2b=-6\]](http://latex.artofproblemsolving.com/6/7/0/6709994caae6e8e9c4f8aa90e2f330c59d4522e3.png)
![\[b=-3\]](http://latex.artofproblemsolving.com/3/f/e/3feb5ee89df7f63099a26a83bdb6f8e95dd1cac2.png)
The answer is 
.
See Also
| 1969 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 6 |
Followed by Problem 8 | |
| 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 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
