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2005 AMC 10A Problems/Problem 10

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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?

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

Solution

A quadratic equation has exactly one root if and only if it is a perfect square. So set

4x^2 + ax + 8x + 9 = (mx + n)^2

4x^2 + ax + 8x + 9 = m^2x^2 + 2mnx + n^2

Two polynomials are equal only if their coefficients are equal, so we must have

m^2 = 4, n^2 = 9

m = \pm 2, n = \pm 3

a + 8= 2mn = \pm 2\cdot 2\cdot 3 = \pm 12

a = 4 or a = -20.

So the desired sum is (4)+(-20)=-16 \Longrightarrow \mathrm{(A)}


Alternatively, note that whatever the two values of a are, they must lead to equations of the form px^2 + qx + r =0 and px^2 - qx + r = 0. So the two choices of a must make a_1 + 8 = q and a_2 + 8 = -q so a_1 + a_2 + 16 = 0 and a_1 + a_2 = -16\Longrightarrow \mathrm{(A)}.

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