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 trinomial has only one root if it is a perfect square.

$(2x\pm3)^2=0$

$4x^2 \pm 12x + 9 = 0$

So the trinomial has only one root when $a+8=\pm12$.

$a = -8\pm12$

$a = 4$ or $a = -20$

So the desired sum is $(4)+(-20)=-16 \Rightarrow A$

See Also

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