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

m (See also)
(Solution to Problem 5)
 
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== 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 ==

Latest revision as of 03:18, 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 (ProblemsAnswer KeyResources)
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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