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Difference between revisions of "1982 AHSME Problems/Problem 27"
(→‎Solution)
 
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<cmath>\begin{align*}
 
<cmath>\begin{align*}
 
c_4(wi)^4+ic_3(wi)^3+c_2(wi)^2+ic_1(wi)+c_0&=0 \\
 
c_4(wi)^4+ic_3(wi)^3+c_2(wi)^2+ic_1(wi)+c_0&=0 \\
−
c_4w^4+c_3w^3-c_2w^2-c_1w+c_0&=0, &&(\bigstar)
+
c_4w^4+c_3w^3-c_2w^2-c_1w+c_0&=0. &&(\bigstar)
 
\end{align*}</cmath>
 
\end{align*}</cmath>
−
which is a polynomial equation in <math>w</math> with real coefficients.
+
Note that <math>(\bigstar)</math> is a polynomial equation in <math>w</math> with real coefficients.
  
 
We are given that <math>w=\frac{a+bi}{i}=b-ai</math> is a solution to <math>(\bigstar).</math> By the Complex Conjugate Root Theorem, we conclude that <math>w=b+ai</math> must also be a solution to <math>(\bigstar),</math> from which <math>z=(b+ai)i=\boxed{\textbf{(C)}\ -a+bi}</math> must also be a solution to the given polynomial equation.
 
We are given that <math>w=\frac{a+bi}{i}=b-ai</math> is a solution to <math>(\bigstar).</math> By the Complex Conjugate Root Theorem, we conclude that <math>w=b+ai</math> must also be a solution to <math>(\bigstar),</math> from which <math>z=(b+ai)i=\boxed{\textbf{(C)}\ -a+bi}</math> must also be a solution to the given polynomial equation.

Latest revision as of 02:51, 7 September 2021

Problem

Suppose $z=a+bi$ is a solution of the polynomial equation \[c_4z^4+ic_3z^3+c_2z^2+ic_1z+c_0=0,\] where $c_0, c_1, c_2, c_3, a,$ and $b$ are real constants and $i^2=-1.$ Which of the following must also be a solution?

$\textbf{(A)}\ -a-bi\qquad \textbf{(B)}\ a-bi\qquad \textbf{(C)}\ -a+bi\qquad \textbf{(D)}\ b+ai \qquad \textbf{(E)}\ \text{none of these}$

Solution

Let $z=wi,$ so the given polynomial equation becomes \begin{align*} c_4(wi)^4+ic_3(wi)^3+c_2(wi)^2+ic_1(wi)+c_0&=0 \\ c_4w^4+c_3w^3-c_2w^2-c_1w+c_0&=0. &&(\bigstar) \end{align*} Note that $(\bigstar)$ is a polynomial equation in $w$ with real coefficients.

We are given that $w=\frac{a+bi}{i}=b-ai$ is a solution to $(\bigstar).$ By the Complex Conjugate Root Theorem, we conclude that $w=b+ai$ must also be a solution to $(\bigstar),$ from which $z=(b+ai)i=\boxed{\textbf{(C)}\ -a+bi}$ must also be a solution to the given polynomial equation.

~MRENTHUSIASM

See Also

1982 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 26 		Followed by
Problem 28
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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