Difference between revisions of "2005 AMC 10A Problems/Problem 10"
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<cmath> \frac{-16 + \sqrt{\text{something}}}{2} + \frac{-16 - \sqrt{\text{something}}}{2}. </cmath> | <cmath> \frac{-16 + \sqrt{\text{something}}}{2} + \frac{-16 - \sqrt{\text{something}}}{2}. </cmath> | ||
<math>\frac{-32}{2} = \boxed{\textbf{(A)}-16}</math> | <math>\frac{-32}{2} = \boxed{\textbf{(A)}-16}</math> | ||
+ | Note: you can also use Vieta's Formulae | ||
== Solution 3== | == Solution 3== |
Latest revision as of 19:33, 13 September 2024
Contents
[hide]Problem
There are two values of for which the equation has only one solution for . What is the sum of those values of ?
Solution 1
A quadratic equation has exactly one root if and only if it is a perfect square. So set
Two polynomials are equal only if their coefficients are equal, so we must have
or .
So the desired sum is
Alternatively, note that whatever the two values of are, they must lead to equations of the form and . So the two choices of must make and so and
Solution 2
Since this quadratic must have a double root, the discriminant of the quadratic formula for this quadratic must be 0. Therefore, we must have We can use the quadratic formula to solve for its roots (we can ignore the expression in the radical sign as it will cancel out due to the sign when added). So we must have Note: you can also use Vieta's Formulae
Solution 3
There is only one positive value for such that the quadratic equation would have only one solution. and are the values of . , so the answer is
Another way of thinking of this is letting the two values of k be and . Since this is obviously a square of a binomial, or for some . Thus, we can say that and . Combining these gives us , so . Our answer is .
~Extremelysupercooldude (Second solution)
See also
2005 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 9 |
Followed by Problem 11 | |
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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