# 1981 AHSME Problems/Problem 30

## Problem

If , , , and are the solutions of the equation , then an equation whose solutions are is

## Solution 1

Using Vieta's formula, we know the sum of the roots is equal to the negative coefficient of the term. Since the coefficient is 0, . Thus, can be rewritten as . Similarly, the other three new roots can be written as , , and .

Now, we need to find a way to transform the function 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, satisfies this criteria.

The new equation, has the required roots and can be simplified to . Since this is not a polynomial, we can multiply both sides by to become . After rearranging and multiplying by negative one, we arrive at so the answer is

## Solution 2

As in solution 1, the roots of the new equation are . Furthermore, applying Vieta’s formula to the original equation yields and . Therefore, the product of the zeros of the new equation is . 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 The function whose roots are the reciprocals of the original equation is therefore . The second term of the chosen equation should satisfy that , hence . The answer is . (Option E looks ridiculous~)