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

Revision as of 20:52, 31 January 2019

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

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 $\boxed{\textbf{(D)} 3x^4-bx^3-1}$

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

Revision as of 20:52, 31 January 2019

Problem

Solution