2013 AMC 10B Problem 19
The real numbers form an arithmetic sequence with . The quadratic has exactly one root. What is this root?
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Key insight
Exactly one root means b^2 = 4ac, and b is halfway between a and c. Take a = 1, solve for c, and the root is -b/(2a).
Solution
1. The numbers form an arithmetic sequence, so is exactly halfway between and :
2. A quadratic has exactly one root when the part under the square root in the quadratic formula is zero: . Then the formula gives the root as .
3. The number cannot be : since , that would make all three , and there would be no quadratic. So .
4. Multiplying , and all by the same positive number does not change the root or either condition. So we may choose to keep things simple. Then , and becomes
5. By the quadratic formula,
6. We need . Since is about , it is too big. So , which is about .
7. Then .
8. The root is
As a sanity check, this is about : negative, as it should be when all the coefficients are positive.
The answer is .
Why this works
Both conditions stay true when , , are all scaled by the same number, so only their ratios matter, and setting leaves just one unknown. The condition is what picks the right sign in step 6. For a quadratic with one root, remember the root is simply .
Alternative approach
Work with the root directly. A quadratic with the single root and is , so and . The arithmetic sequence condition becomes , or , so . We also need , that is , so . Only passes.
Concise solution
A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.
An arithmetic sequence means is the average: . Exactly one root means the discriminant vanishes: . Substituting,
Here (otherwise everything is and there is no quadratic), so divide by and set :
Since , we need , so .
The double root of is . With ,
The answer is .
Review in the AoPS books
- Introduction to Algebra, Chapter 10: Quadratic Equations – Part 1 (§10.2, §10.3, §10.4)
- Introduction to Algebra, Chapter 21: Sequences & Series (§21.1, §21.2, §21.3, §21.4)
- Introduction to Algebra, Chapter 11: Special Factorizations
- Intermediate Algebra, Chapter 20: Some Advanced Strategies (§20.2)
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Problem © Mathematical Association of America (MAA), American Mathematics Competitions. Reproduced for non-commercial educational use. Solution and commentary are original to this site.