1966 AHSME Problems/Problem 30

Problem

If three of the roots of $x^4+ax^2+bx+c=0$ are $1$ , $2$ , and $3$ then the value of $a+c$ is:

$\text{(A) } 35 \quad \text{(B) } 24 \quad \text{(C) } -12 \quad \text{(D) } -61 \quad \text{(E) } -63$

Solution

Since this is a quartic equation, there are going to be $4$ solutions. By Vieta's formulas, since the $x^3$ term is $0$ , the sum of the roots is also $0$ . Therefore, the 4th root of this polynomial is $-6$ . Lastly, by Vieta's Formulas, $a+c=-(1\cdot2+1\cdot3+1\cdot-6+2\cdot3+2\cdot-6+3\cdot-6)+1\cdot2\cdot3\cdot-6=-25-36=-61= \fbox{D}$

1966 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 29	Followed by Problem 31
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
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1966 AHSME Problems/Problem 30

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