Difference between revisions of "1952 AHSME Problems/Problem 23"
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==Solution== | ==Solution== | ||
| − | Cross-multiplying, we find that <math> (m+1)x^2-(bm+am+b-a)x+c(m-1)=0 </math>. Because the roots of this quadratic are | + | Cross-multiplying, we find that <math> (m+1)x^2-(bm+am+b-a)x+c(m-1)=0 </math>. Because the roots of this quadratic are additive inverses, their sum is <math> 0 </math>. According to [[Vieta's Formulas]], <math> \frac{bm+am+b-a}{m+1}=0 </math>, or <math> m(a+b)=a-b </math>. Hence, <math> m=\boxed{\textbf{(A)}\ \frac{a-b}{a+b}} </math>. |
==See also== | ==See also== | ||
{{AHSME 50p box|year=1952|num-b=22|num-a=24}} | {{AHSME 50p box|year=1952|num-b=22|num-a=24}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Latest revision as of 22:59, 28 January 2014
Problem
If 
has roots which are numerically equal but of opposite signs, the value of 
must be:
Solution
Cross-multiplying, we find that 
. Because the roots of this quadratic are additive inverses, their sum is 
. According to Vieta's Formulas, 
, or 
. Hence, 
.
See also
| 1952 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 22 |
Followed by Problem 24 | |
| 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 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
