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Difference between revisions of "2005 AMC 10A Problems/Problem 10"
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==Solution==
 
==Solution==
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A [[quadratic equation]] has exactly one root if and only if it is a [[perfect square]].  So set
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A [[quadratic equation]] has exactly one [[root]] if and only if it is a [[perfect square]].  So set
 
<math>4x^2 + ax + 8x + 9 = (mx + n)^2</math>
 
<math>4x^2 + ax + 8x + 9 = (mx + n)^2</math>
 
<math>4x^2 + ax + 8x + 9 = m^2x^2 + 2mnx + n^2</math>
 
<math>4x^2 + ax + 8x + 9 = m^2x^2 + 2mnx + n^2</math>

Revision as of 09:49, 2 August 2006

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

See Also

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