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> | ||
− | == Solution == | + | == 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> | ||
+ | |||
+ | == Solution 2 == | ||
+ | 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> | ||
+ | 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}</math>. The second term of the chosen equation <math>a</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>. | ||
+ | (Option E looks ridiculous~) |
Revision as of 20:22, 9 October 2020
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~)