Difference between revisions of "2013 AMC 10B Problems/Problem 19"
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==Problem== | ==Problem== | ||
− | The real numbers <math>c,b,a</math> form an arithmetic sequence with <math>a\ge b\ge c | + | The real numbers <math>c,b,a</math> form an arithmetic sequence with <math>a\ge b\ge c> 0</math>. The quadratic <math>ax^2+bx+c</math> has exactly one root. What is this root? |
<math> \textbf{(A)}\ -7-4\sqrt{3}\qquad\textbf{(B)}\ -2-\sqrt{3}\qquad\textbf{(C)}\ -1\qquad\textbf{(D)}\ -2+\sqrt{3}\qquad\textbf{(E)}\ -7+4\sqrt{3} </math> | <math> \textbf{(A)}\ -7-4\sqrt{3}\qquad\textbf{(B)}\ -2-\sqrt{3}\qquad\textbf{(C)}\ -1\qquad\textbf{(D)}\ -2+\sqrt{3}\qquad\textbf{(E)}\ -7+4\sqrt{3} </math> |
Revision as of 15:14, 28 August 2019
Contents
Problem
The real numbers form an arithmetic sequence with . The quadratic has exactly one root. What is this root?
Solutions
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 can get . Ignoring the negatives(for now), 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 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
We have Hence .
And we have . Squaring the first expression, and dividing by 4, we get .
Setting the two equations equal, we have .
.
Dividing by , we get .
Setting we have
Solving for , we get We know that , so .
By Vieta's, we know that , where equals the double root of the quadratic. So, we get . After rationalizing the denominator, we get .
Hence, . Solving for , we have equals .
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 |
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