1981 AHSME Problems/Problem 30


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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