Difference between revisions of "1968 AHSME Problems/Problem 13"
(Created page with "== Problem == If <math>m</math> and <math>n</math> are the roots of <math>x^2+mx+n=0 ,m \ne 0,n \ne 0</math>, then the sum of the roots is: <math>\text{(A) } -\frac{1}{2}\quad ...") |
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== Solution == | == Solution == | ||
| − | <math>\fbox{}</math> | + | By Vieta's Theorem, <math>mn = n</math> and <math>-(m + n) = m</math>. Dividing the first equation by <math>n</math> gives <math>m = 1</math>. Multiplying the 2nd by -1 gives <math>m + n = -m</math>. The RHS is -1, so the answer is <math>\fbox{B}</math> |
== See also == | == See also == | ||
| − | {{AHSME box|year=1968|num-b=12|num-a=14}} | + | {{AHSME 35p box|year=1968|num-b=12|num-a=14}} |
[[Category: Introductory Algebra Problems]] | [[Category: Introductory Algebra Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Latest revision as of 01:52, 16 August 2023
Problem
If 
and 
are the roots of 
, then the sum of the roots is:
Solution
By Vieta's Theorem, 
and 
. Dividing the first equation by gives 
. Multiplying the 2nd by -1 gives 
. The RHS is -1, so the answer is 
See also
| 1968 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 12 |
Followed by Problem 14 | |
| 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. 
