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Difference between revisions of "1969 AHSME Problems/Problem 5"

(Created page with "== Problem == If a number <math>N,N \ne 0</math>, diminished by four times its reciprocal, equals a given real constant <math>R</math>, then, for this given <math>R</math>, the s...")
 
(Solution to Problem 5)
 
(One intermediate revision by one other user not shown)
Line 9: Line 9:
  
 
== Solution ==
 
== Solution ==
−
<math>\fbox{B}</math>
+
Write an equation from the given information.
 +
<cmath>N - \frac{4}{N} = R</cmath>
 +
<cmath>N^2 - 4 = RN</cmath>
 +
<cmath>N^2 - RN - 4 = 0</cmath>
 +
By [[Vieta's Formulas]], the sum of all possible values of <math>N</math> for a given <math>R</math> is <math>R</math>, so the answer is <math>\boxed{\textbf{(B)}}</math>.
  
 
== See also ==
 
== See also ==
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{{AHSME box|year=1969|num-b=4|num-a=6}}   
+
{{AHSME 35p box|year=1969|num-b=4|num-a=6}}   
  
 
[[Category: Introductory Algebra Problems]]
 
[[Category: Introductory Algebra Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 02:18, 7 June 2018

Problem

If a number $N,N \ne 0$, diminished by four times its reciprocal, equals a given real constant $R$, then, for this given $R$, the sum of all such possible values of $N$ is

$\text{(A) } \frac{1}{R}\quad \text{(B) } R\quad \text{(C) } 4\quad \text{(D) } \frac{1}{4}\quad \text{(E) } -R$

Solution

Write an equation from the given information. \[N - \frac{4}{N} = R\] \[N^2 - 4 = RN\] \[N^2 - RN - 4 = 0\] By Vieta's Formulas, the sum of all possible values of $N$ for a given $R$ is $R$, so the answer is $\boxed{\textbf{(B)}}$.

See also

1969 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 4 		Followed by
Problem 6
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

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