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Difference between revisions of "1995 AIME Problems/Problem 5"

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== Problem ==
 
== Problem ==
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For certain real values of <math>\displaystyle a, b, c,</math> and <math>\displaystyle d_{},</math>  the equation <math>\displaystyle x^4+ax^3+bx^2+cx+d=0</math> has four non-real roots.  The product of two of these roots is <math>\displaystyle 13+i</math> and the sum of the other two roots is <math>\displaystyle 3+4i,</math> where <math>i=\sqrt{-1}.</math>  Find <math>\displaystyle b.</math>
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For certain real values of <math>a, b, c,</math> and <math>d_{},</math>  the equation <math>x^4+ax^3+bx^2+cx+d=0</math> has four non-real roots.  The product of two of these roots is <math>13+i</math> and the sum of the other two roots is <math>3+4i,</math> where <math>i=\sqrt{-1}.</math>  Find <math>b.</math>
  
 
== Solution ==
 
== Solution ==
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Since the [[coefficient]]s of the [[polynomial]] are real, it follows that the non-real roots must come in [[complex conjugate]] pairs. Let the first two roots be <math>m,n</math>. Since <math>m+n</math> is not real, <math>m,n</math> are not conjugates, so the other pair of roots must be the conjugates of <math>m,n</math>. Let <math>m'</math> be the conjugate of <math>m</math>, and <math>n'</math> be the conjugate of <math>n</math>. Then,
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<cmath>m\cdot n = 13 + i,m' + n' = 3 + 4i\Longrightarrow m'\cdot n' = 13 - i,m + n = 3 - 4i.</cmath>
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By [[Vieta's formulas]], we have that <math>b = mm' + nn' + mn' + nm' + mn + m'n' = (m + n)(m' + n') + mn + m'n' = \boxed{051}</math>.
  
 
== See also ==
 
== See also ==
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* [[1995_AIME_Problems/Problem_4|Previous Problem]]
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{{AIME box|year=1995|num-b=4|num-a=6|t=394478}}
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* [[1995_AIME_Problems/Problem_6|Next Problem]]
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* [[1995 AIME Problems]]
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[[Category:Intermediate Algebra Problems]]

Revision as of 11:12, 18 June 2008

Problem

For certain real values of $a, b, c,$ and $d_{},$ the equation $x^4+ax^3+bx^2+cx+d=0$ has four non-real roots. The product of two of these roots is $13+i$ and the sum of the other two roots is $3+4i,$ where $i=\sqrt{-1}.$ Find $b.$

Solution

Since the coefficients of the polynomial are real, it follows that the non-real roots must come in complex conjugate pairs. Let the first two roots be $m,n$. Since $m+n$ is not real, $m,n$ are not conjugates, so the other pair of roots must be the conjugates of $m,n$. Let $m'$ be the conjugate of $m$, and $n'$ be the conjugate of $n$. Then, \[m\cdot n = 13 + i,m' + n' = 3 + 4i\Longrightarrow m'\cdot n' = 13 - i,m + n = 3 - 4i.\] By Vieta's formulas, we have that $b = mm' + nn' + mn' + nm' + mn + m'n' = (m + n)(m' + n') + mn + m'n' = \boxed{051}$.

See also

1995 AIME (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
All AIME Problems and Solutions
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