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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>
  
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== Solution ==
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== Solution 1 ==
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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>.
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== Solution 2 ==
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Let's assume that the 2 roots multiplied together are p+qi, and r+si, and the two roots added together are the conjugates of the previous roots. Using Vieta, we get
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<cmath>b = (p+qi)(r+si) + (p+qi)(r-si) + (p-qi)(r+si) + (p-qi)(r-si) + (p+qi)(p-qi) + (r+si)(r-si) =</cmath>
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<cmath>(p+qi+p-qi)(r+si+r-si) + (p+qi)(p-qi) + (r+si)(r-si) =</cmath>
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<cmath>(2p)(2r) + p^2 + q^2 + r^2 + s^2 = 4pr + p^2 + q^2 + r^2 + s^2 = (p+r)^2 + 2pr + q^2 + s^2</cmath>
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We are now stuck. We can't simplify further. But, we look back to the problem and see that the two roots that are multiplied together give a product of <math>13+i</math>, and the two roots that are added give <math>3+4i</math>. This gets three equations necessary for solving the problem. <cmath>p+r = 3</cmath> <cmath>pr-qs = 13</cmath> <cmath>-q-s = 4</cmath> So, alright. Let's use the first equation to get that <math>(p+r)^2 = 9</math>, and substitute that in.  Now, the equation becomes:
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<cmath>b = 9 + 2pr + q^2 + s^2</cmath>
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We wish that we can turn the 2pr into 2qs. Fortunately, we can do that. By using the second equation, we can manipulate it to be <math>pr = 13+qs</math>, and substitute that in.
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<cmath>b = 9 + 2(13+qs) + q^2 + s^2 = 9 + 26 + 2qs + q^2 + s^2 = 35 + (q+s)^2</cmath>
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We can square both sides of the third equation, and get <math>(q+s)^2 = 16</math> We substitute that in and we get
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<cmath>b = 35+16 = \boxed{051}</cmath>
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- AlexLikeMath
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- Corrections by VSPuzzler
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- small correction be Marshall_Huang
  
 
== 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}}
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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]]
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{{MAA Notice}}

Latest revision as of 13:08, 4 July 2024

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 1

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 $b = mm' + nn' + mn' + nm' + mn + m'n' = (m + n)(m' + n') + mn + m'n' = \boxed{051}$.

Solution 2

Let's assume that the 2 roots multiplied together are p+qi, and r+si, and the two roots added together are the conjugates of the previous roots. Using Vieta, we get \[b = (p+qi)(r+si) + (p+qi)(r-si) + (p-qi)(r+si) + (p-qi)(r-si) + (p+qi)(p-qi) + (r+si)(r-si) =\]

\[(p+qi+p-qi)(r+si+r-si) + (p+qi)(p-qi) + (r+si)(r-si) =\]

\[(2p)(2r) + p^2 + q^2 + r^2 + s^2 = 4pr + p^2 + q^2 + r^2 + s^2 = (p+r)^2 + 2pr + q^2 + s^2\]

We are now stuck. We can't simplify further. But, we look back to the problem and see that the two roots that are multiplied together give a product of $13+i$, and the two roots that are added give $3+4i$. This gets three equations necessary for solving the problem. \[p+r = 3\] \[pr-qs = 13\] \[-q-s = 4\] So, alright. Let's use the first equation to get that $(p+r)^2 = 9$, and substitute that in. Now, the equation becomes:

\[b = 9 + 2pr + q^2 + s^2\]

We wish that we can turn the 2pr into 2qs. Fortunately, we can do that. By using the second equation, we can manipulate it to be $pr = 13+qs$, and substitute that in.

\[b = 9 + 2(13+qs) + q^2 + s^2 = 9 + 26 + 2qs + q^2 + s^2 = 35 + (q+s)^2\]

We can square both sides of the third equation, and get $(q+s)^2 = 16$ We substitute that in and we get

\[b = 35+16 = \boxed{051}\]

- AlexLikeMath

- Corrections by VSPuzzler

- small correction be Marshall_Huang

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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