Difference between revisions of "1966 AHSME Problems/Problem 30"
(Created page with "== Problem == If three of the roots of <math>x^4+ax^2+bx+c=0</math> are <math>1</math>, <math>2</math>, and <math>3</math> then the value of <math>a+c</math> is: <math>\text{(A)...") |
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== Solution == | == Solution == | ||
| − | + | Since this is a quartic equation, there are going to be <math>4</math> solutions. | |
| + | By Vieta's formulas, since the <math>x^3</math> term is <math>0</math>, the sum of the roots is also <math>0</math>. Therefore, the 4th root of this polynomial is <math>-6</math>. | ||
| + | Lastly, by Vieta's Formulas, <math>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}</math> | ||
== See also == | == See also == | ||
Latest revision as of 14:38, 26 July 2021
Problem
If three of the roots of 
are 
, 
, and 
then the value of 
is:
Solution
Since this is a quartic equation, there are going to be 
solutions.
By Vieta's formulas, since the 
term is 
, the sum of the roots is also . Therefore, the 4th root of this polynomial is 
.
Lastly, by Vieta's Formulas, 
See also
| 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 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
