Difference between revisions of "1968 AHSME Problems/Problem 13"

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== Solution ==
 
== Solution ==
<math>\fbox{B}</math>
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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 ==

Latest revision as of 14:49, 25 November 2020

Problem

If $m$ and $n$ are the roots of $x^2+mx+n=0 ,m \ne 0,n \ne 0$, then the sum of the roots is:

$\text{(A) } -\frac{1}{2}\quad \text{(B) } -1\quad \text{(C) } \frac{1}{2}\quad \text{(D) } 1\quad \text{(E) } \text{undetermined}$

Solution

By Vieta's Theorem, $mn = n$ and $-(m + n) = m$. Dividing the first equation by $n$ gives $m = 1$. Multiplying the 2nd by -1 gives $m + n = -m$. The RHS is -1, so the answer is $\fbox{B}$

See also

1968 AHSME (ProblemsAnswer KeyResources)
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
All AHSME Problems and Solutions

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