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 $a$ ?

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

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 $(4)+(-20)=-16 \Longrightarrow \mathrm{(A)}$

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2005 AMC 10A Problems/Problem 10

Problem

Solution

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