Difference between revisions of "2005 AMC 10A Problems/Problem 10"

(added problem and solution)
 
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==Solution==
 
==Solution==
A [[trinomial]] has only one root if it is a [[perfect square]].  
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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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<math>4x^2 + ax + 8x + 9 = (mx + n)^2</math>
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<math>4x^2 + ax + 8x + 9 = m^2x^2 + 2mnx + n^2</math>
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Two [[polynomial]]s are equal only if their [[coefficient]]s are equal, so we must have
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<math>m^2 = 4, n^2 = 9</math>
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<math>m = \pm 2, n = \pm 3</math>
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<math>a + 8= 2mn = \pm 2\cdot 2\cdot 3 = \pm 12</math>
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<math>a = 4</math> or <math>a = -20</math>.
  
<math>(2x\pm3)^2=0</math>
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So the desired sum is <math> (4)+(-20)=-16 \Longrightarrow \mathrm{(A)} </math>
 
 
<math> 4x^2 \pm 12x + 9 = 0 </math>
 
 
 
So the trinomial has only one root when <math>a+8=\pm12</math>.
 
 
 
<math> a = -8\pm12 </math>
 
 
 
<math> a = 4 </math> or <math> a = -20 </math>
 
 
 
So the desired sum is <math> (4)+(-20)=-16 \Rightarrow A </math>
 
 
   
 
   
 
==See Also==
 
==See Also==

Revision as of 09:48, 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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