Difference between revisions of "2013 AMC 10B Problems/Problem 19"
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Let the double root be <math>r</math>. Then by the arithmetic progression and Vieta's,<cmath>\begin{aligned}a-b & =b-c\ | Let the double root be <math>r</math>. Then by the arithmetic progression and Vieta's,<cmath>\begin{aligned}a-b & =b-c\ | ||
1-\frac{b}a & =\frac{b}a-\frac{c}a\ | 1-\frac{b}a & =\frac{b}a-\frac{c}a\ |
Revision as of 16:12, 2 February 2014
Contents
[hide]Problem
The real numbers form an arithmetic sequence with
The quadratic
has exactly one root. What is this root?
Solution
Solution 1
It is given that has 1 real root, so the discriminant is zero, or
. Because a, b, c are in arithmetic progression,
, or
. We need to find the unique root, or
(discriminant is 0). From
, we have
. Ignoring the negatives, we have
. Fortunately, finding
is not very hard. Plug in
to
, we have
, or
, and dividing by
gives
, so
. But
, violating the assumption that
. Therefore,
. Plugging this in, we have
. But we need the negative of this, so the answer is
.
Solution 2
Note that we can divide the polynomial by to make the leading coefficient 1 since dividing does not change the roots or the fact that the coefficients are in an arithmetic sequence. Also, we know that there is exactly one root so this equation is must be of the form
where
.
We now use the fact that the coefficients are in an arithmetic sequence. Note that in any arithmetic sequence, the average is equal to the median. Thus,
and
. Since
, we easily see that
has to be between 1 and 0. Thus, we can eliminate
and are left with
as the answer.
Solution 3
Given that has only 1 real root, we know that the discriminant must equal 0, or that
. Because the discriminant equals 0, we have that the root of the quadratic is
. We are also given that the coefficients of the quadratic are in arithmetic progression, where
. Letting the arbitrary difference equal variable
, we have that
and that
. Plugging those two equations into
, we have
which yields
. Isolating
, we have
. Substituting that in for
in
, we get
. Once again, substituting that in for
in
, we have
. The answer is
.
Solution 4
Let the double root be . Then by the arithmetic progression and Vieta's,
\begin{aligned}a-b & =b-c\\ 1-\frac{b}a & =\frac{b}a-\frac{c}a\\ 1+2r & =-2r-r^2\\ r^2+4r+1 & =0\\ r & =-2\pm\sqrt{3}\end{aligned} (Error compiling LaTeX. Unknown error_msg)
We see , and so we want
. Note that since
and
, we can conclude that
and the answer is
.
See also
2013 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 18 |
Followed by Problem 20 | |
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.