Difference between revisions of "2005 AMC 12A Problems/Problem 9"
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== Solution == | == Solution == | ||
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A [[quadratic equation]] always has two roots, unless it has a double root. That means we can write the quadratic as a square, and the coefficients 4 and 9 suggest this. [[Completing the square]], <math>0 = (2x \pm 3)^2 = 4x^2 \pm 12x + 9</math>, so <math>\pm 12 = a + 8 \Longrightarrow a = 4, -20</math>. The sum of these is <math>-20 + 4 = -16 \Rightarrow \mathrm{(A)}</math>. | A [[quadratic equation]] always has two roots, unless it has a double root. That means we can write the quadratic as a square, and the coefficients 4 and 9 suggest this. [[Completing the square]], <math>0 = (2x \pm 3)^2 = 4x^2 \pm 12x + 9</math>, so <math>\pm 12 = a + 8 \Longrightarrow a = 4, -20</math>. The sum of these is <math>-20 + 4 = -16 \Rightarrow \mathrm{(A)}</math>. | ||
| + | ==Solution 2 === | ||
Another method would be to use the quadratic formula, since our <math>x^2</math> coefficient is given as 4, the <math>x</math> coefficient is <math>a+8</math> and the constant term is <math>9</math>. Hence, <math>x = \frac{-(a+8) \pm \sqrt {(a+8)^2-4(4)(9)}}{2(4)}</math> Because we want only a single solution for <math>x</math>, the determinant must equal 0. Therefore, we can write <math>(a+8)^2 - 144 = 0</math> which factors to <math>a^2 + 16a - 80 = 0</math>; using [[Vieta's formulas]] we see that the sum of the solutions for <math>a</math> is the opposite of the coefficient of <math>a</math>, or <math>-16 \Rightarrow \mathrm{ (A)}</math>. | Another method would be to use the quadratic formula, since our <math>x^2</math> coefficient is given as 4, the <math>x</math> coefficient is <math>a+8</math> and the constant term is <math>9</math>. Hence, <math>x = \frac{-(a+8) \pm \sqrt {(a+8)^2-4(4)(9)}}{2(4)}</math> Because we want only a single solution for <math>x</math>, the determinant must equal 0. Therefore, we can write <math>(a+8)^2 - 144 = 0</math> which factors to <math>a^2 + 16a - 80 = 0</math>; using [[Vieta's formulas]] we see that the sum of the solutions for <math>a</math> is the opposite of the coefficient of <math>a</math>, or <math>-16 \Rightarrow \mathrm{ (A)}</math>. | ||
Revision as of 08:39, 1 July 2008
Problem
There are two values of 
for which the equation 
has only one solution for 
. What is the sum of these values of ?
Solution
Solution 1
A quadratic equation always has two roots, unless it has a double root. That means we can write the quadratic as a square, and the coefficients 4 and 9 suggest this. Completing the square, 
, so 
. The sum of these is 
.
Solution 2 =
Another method would be to use the quadratic formula, since our 
coefficient is given as 4, the coefficient is 
and the constant term is 
. Hence, 
Because we want only a single solution for , the determinant must equal 0. Therefore, we can write 
which factors to 
; using Vieta's formulas we see that the sum of the solutions for is the opposite of the coefficient of , or 
.
See also
| 2005 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by Problem 8 |
Followed by Problem 10 |
| 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 12 Problems and Solutions | |
