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2005 AMC 10A Problems/Problem 10

Problem

There are two values of $a$ for which the equation $4x^2 + ax + 8x + 9 = 0$ has only one solution for $x$. What is the sum of those values of $a$?

$\textbf{(A) } -16 \qquad\textbf{(B) } -8 \qquad\textbf{(C) } 0 \qquad\textbf{(D) } 8 \qquad\textbf{(E) } 20$

Solution 1

A quadratic equation has exactly $1$ distinct root if and only if the left-hand side is a perfect square. So we require

\[4x^2 + ax + 8x + 9 = (mx + n)^2 = m^2 x^2 + 2mnx + n^2.\]

Two polynomials are equal only if their coefficients are equal, so we must have $m^2 = 4$ and $n^2 = 9$, which reduce to $m = \pm 2$ and $n = \pm 3$ respectively. By equating $x$-coefficients, it follows that $a + 8 = 2mn = \pm 2 \cdot 2\cdot 3 = \pm 12$, so either $a = 12-8 = 4$ or $a = -12-8 = -20$.

Accordingly, the desired sum is $4+(-20) = \boxed{\textbf{(A) } -16}$.

(Alternatively, we can observe that whatever the $2$ values of $a$ are, they must lead to equations of the form $px^2 + qx + r = 0$ and $px^2 - qx + r = 0$, corresponding to the perfect squares $\left(mx \pm n\right)^2 = 0$. So the $2$ choices of $a$, say $a_1$ and $a_2$, must satisfy $a_1 + 8 = q$ and $a_2 + 8 = -q$, and adding these equations yields $a_1 + a_2 + 16 = 0 \iff a_1 + a_2 = \boxed{\textbf{(A) } -16}$.)

Solution 2

Since this quadratic must have a double root, its discriminant must be $0$. Therefore we require \[(a+8)^2 - 4 \cdot 4 \cdot 9 = 0 \iff a^2 + 16a - 80 = 0.\] At this point, Vieta's formulae directly give the sum of the $2$ possible values of $a$ as $-\frac{16}{1} = \boxed{\textbf{(A) } -16}.$

(Alternatively, we could use the quadratic formula to directly solve this equation for $a$; the precise value of the expression under the square root does not actually matter, as it will cancel out due to the $\pm$ signs when added. This again gives the required sum as \[\frac{-16 + \sqrt{\text{something}}}{2} + \frac{-16 - \sqrt{\text{something}}}{2} = \frac{-32}{2} = \boxed{\textbf{(A) } -16}.)\]

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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