Difference between revisions of "2003 AMC 10A Problems/Problem 18"
Mathboy282 (talk | contribs) (→Solution 2) |
Pi is 3.14 (talk | contribs) (→Solution 1) |
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<math> \mathrm{(A) \ } -\frac{2004}{2003}\qquad \mathrm{(B) \ } -1\qquad \mathrm{(C) \ } \frac{2003}{2004}\qquad \mathrm{(D) \ } 1\qquad \mathrm{(E) \ } \frac{2004}{2003} </math> | <math> \mathrm{(A) \ } -\frac{2004}{2003}\qquad \mathrm{(B) \ } -1\qquad \mathrm{(C) \ } \frac{2003}{2004}\qquad \mathrm{(D) \ } 1\qquad \mathrm{(E) \ } \frac{2004}{2003} </math> | ||
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+ | == Video Solution == | ||
+ | https://youtu.be/3dfbWzOfJAI?t=456 | ||
+ | |||
+ | ~ pi_is_3.14 | ||
== Solution 1 == | == Solution 1 == |
Revision as of 02:07, 14 January 2021
Contents
[hide]Problem
What is the sum of the reciprocals of the roots of the equation ?
Video Solution
https://youtu.be/3dfbWzOfJAI?t=456
~ pi_is_3.14
Solution 1
Multiplying both sides by :
Let the roots be and .
The problem is asking for
By Vieta's formulas:
So the answer is .
Solution 2
Dividing both sides by ,
,
we see by Vieta's formulas that the sum of the roots is .
Solution 3(Alternate Approach to Solution 1)
Re-stating and multiplying by , we have .
Simplifying, we have
Putting this together, we have
Simplifying, we have
Multiplying both sides by we have
Moving to the left side, we have
Let's go back and see what we want.
We want the sum of the reciprocals of the roots, in which if the roots were to be called and we would want .
We can use Vieta's formulas to solve the sum of the roots and the product of the roots.
We have that the sum of the two roots is where the quadratic is
In this case, and .
Therefore,
We have that the product of the two roots is where the quadratic is
In this case, and
Therefore,
Now that we have the sum and product of the roots, we can substitute back into the expression of what we want.
We have
~mathboy282
See Also
2003 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 17 |
Followed by Problem 19 | |
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.