Difference between revisions of "2006 AMC 10B Problems/Problem 14"
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== Solution == | == Solution == | ||
| − | In a [[quadratic equation]] in the form <math> x^2 + bx + c = 0 </math>, the product of the [[root]]s is <math>c</math>. | + | In a [[quadratic equation]] in the form <math> x^2 + bx + c = 0 </math>, the product of the [[root]]s is <math>c</math>(Vieta's Formulas). |
Using this property, we have that <math>ab=2</math> and | Using this property, we have that <math>ab=2</math> and | ||
<math> q = (a+\frac{1}{b})\cdot(b+\frac{1}{a}) = \frac{ab+1}{b} \cdot \frac{ab+1}{a} = \frac{(ab+1)^2}{ab} = \frac{(2+1)^2}{2} = \frac{9}{2} \Rightarrow D </math> | <math> q = (a+\frac{1}{b})\cdot(b+\frac{1}{a}) = \frac{ab+1}{b} \cdot \frac{ab+1}{a} = \frac{(ab+1)^2}{ab} = \frac{(2+1)^2}{2} = \frac{9}{2} \Rightarrow D </math> | ||
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| + | *Notice the fact that we never actually found the roots. | ||
== See Also == | == See Also == | ||
Revision as of 20:35, 1 February 2014
Problem
Let 
and 
be the roots of the equation 
. Suppose that 
and 
are the roots of the equation 
. What is 
?
Solution
In a quadratic equation in the form 
, the product of the roots is 
(Vieta's Formulas).
Using this property, we have that 
and
- Notice the fact that we never actually found the roots.
See Also
| 2006 AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by Problem 13 |
Followed by Problem 15 | |
| 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 | ||
| All AMC 10 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
