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Difference between revisions of "1981 AHSME Problems/Problem 30"

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<math> \textbf{(A)}\ 3x^4 + bx + 1 = 0\qquad \textbf{(B)}\ 3x^4 - bx + 1 = 0\qquad \textbf{(C)}\ 3x^4 + bx^3 - 1 = 0\qquad \\\textbf{(D)}\ 3x^4 - bx^3 - 1 = 0\qquad  \textbf{(E)}\ \text{none of these}</math>
 
<math> \textbf{(A)}\ 3x^4 + bx + 1 = 0\qquad \textbf{(B)}\ 3x^4 - bx + 1 = 0\qquad \textbf{(C)}\ 3x^4 + bx^3 - 1 = 0\qquad \\\textbf{(D)}\ 3x^4 - bx^3 - 1 = 0\qquad  \textbf{(E)}\ \text{none of these}</math>
  
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== Solution ==
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== Solution 1 ==
 
Using Vieta's formula, we know the sum of the roots is equal to the negative coefficient of the <math>x^3</math> term. Since the coefficient is 0, <math>a+b+c+d=0</math>. Thus, <math>\frac{a+b+c}{d^2}</math> can be rewritten as <math>\frac{-d}{d^2}=\frac{1}{-d}</math>. Similarly, the other three new roots can be written as <math>\frac{1}{-c}</math>, <math>\frac{1}{-b}</math>, and <math>\frac{1}{-a}</math>.
 
Using Vieta's formula, we know the sum of the roots is equal to the negative coefficient of the <math>x^3</math> term. Since the coefficient is 0, <math>a+b+c+d=0</math>. Thus, <math>\frac{a+b+c}{d^2}</math> can be rewritten as <math>\frac{-d}{d^2}=\frac{1}{-d}</math>. Similarly, the other three new roots can be written as <math>\frac{1}{-c}</math>, <math>\frac{1}{-b}</math>, and <math>\frac{1}{-a}</math>.
  
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The new equation, <math>f(\frac{1}{-x})=0</math> has the required roots and can be simplified to <math>\frac{1}{x^4}+\frac{b}{x}-3=0</math>. Since this is not a polynomial, we can multiply both sides by <math>x^4</math> to become <math>1+bx^3-3x^4=0</math>. After rearranging and multiplying by negative one, we arrive at <math>3x^4-bx^3-1</math> so the answer is <math>\boxed{\textbf{(D)} 3x^4-bx^3-1}</math>
 
The new equation, <math>f(\frac{1}{-x})=0</math> has the required roots and can be simplified to <math>\frac{1}{x^4}+\frac{b}{x}-3=0</math>. Since this is not a polynomial, we can multiply both sides by <math>x^4</math> to become <math>1+bx^3-3x^4=0</math>. After rearranging and multiplying by negative one, we arrive at <math>3x^4-bx^3-1</math> so the answer is <math>\boxed{\textbf{(D)} 3x^4-bx^3-1}</math>
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== Solution 2 ==
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As in solution 1, the roots of the new equation are <math>-\frac{1}{a}, -\frac{1}{b}, -\frac{1}{c},-\frac{1}{d}</math>. Furthermore, applying Vieta’s formula to the original equation yields <math>abcd=-3</math> and <math>abc+abd+acd+bcd=-3(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d})=b</math>. Therefore, the product of the zeros of the new equation is <math>\frac{1}{abcd}=-\frac{1}{3}</math>. This limits our choices to options C and D, and we need to look for the sum of the roots of the new equation. This sum equals to <math>-(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d})</math>
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The function whose roots are the reciprocals of the original equation is <math>-3x^4-bx^3+1</math> therefore <math>-(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d})=-(-\frac{-b}{-3})=\frac{b}{3}</math>. The second term of the chosen equation <math>a_{2}</math> should satisfy that <math>-\frac{a_{2}}{3}=\frac{b}{3}</math>, hence <math>a_{2}=-b</math>. The answer is <math>D</math>.
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(Option E looks ridiculous~)

Latest revision as of 20:31, 9 October 2020

Problem

If $a$, $b$, $c$, and $d$ are the solutions of the equation $x^4 - bx - 3 = 0$, then an equation whose solutions are \[\dfrac {a + b + c}{d^2}, \dfrac {a + b + d}{c^2}, \dfrac {a + c + d}{b^2}, \dfrac {b + c + d}{a^2}\]is

$\textbf{(A)}\ 3x^4 + bx + 1 = 0\qquad \textbf{(B)}\ 3x^4 - bx + 1 = 0\qquad \textbf{(C)}\ 3x^4 + bx^3 - 1 = 0\qquad \\\textbf{(D)}\ 3x^4 - bx^3 - 1 = 0\qquad \textbf{(E)}\ \text{none of these}$

Solution 1

Using Vieta's formula, we know the sum of the roots is equal to the negative coefficient of the $x^3$ term. Since the coefficient is 0, $a+b+c+d=0$. Thus, $\frac{a+b+c}{d^2}$ can be rewritten as $\frac{-d}{d^2}=\frac{1}{-d}$. Similarly, the other three new roots can be written as $\frac{1}{-c}$, $\frac{1}{-b}$, and $\frac{1}{-a}$.

Now, we need to find a way to transform the function $f(x)=x^4-bx-3$ such that all the roots are its negative reciprocal. We can create this new function by taking the negative reciprocal of the argument. In other words, $f(\frac{1}{-x})$ satisfies this criteria.

The new equation, $f(\frac{1}{-x})=0$ has the required roots and can be simplified to $\frac{1}{x^4}+\frac{b}{x}-3=0$. Since this is not a polynomial, we can multiply both sides by $x^4$ to become $1+bx^3-3x^4=0$. After rearranging and multiplying by negative one, we arrive at $3x^4-bx^3-1$ so the answer is $\boxed{\textbf{(D)} 3x^4-bx^3-1}$

Solution 2

As in solution 1, the roots of the new equation are $-\frac{1}{a}, -\frac{1}{b}, -\frac{1}{c},-\frac{1}{d}$. Furthermore, applying Vieta’s formula to the original equation yields $abcd=-3$ and $abc+abd+acd+bcd=-3(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d})=b$. Therefore, the product of the zeros of the new equation is $\frac{1}{abcd}=-\frac{1}{3}$. This limits our choices to options C and D, and we need to look for the sum of the roots of the new equation. This sum equals to $-(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d})$ The function whose roots are the reciprocals of the original equation is $-3x^4-bx^3+1$ therefore $-(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d})=-(-\frac{-b}{-3})=\frac{b}{3}$. The second term of the chosen equation $a_{2}$ should satisfy that $-\frac{a_{2}}{3}=\frac{b}{3}$, hence $a_{2}=-b$. The answer is $D$. (Option E looks ridiculous~)

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