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

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<math> \mathrm{(A) \ } -16\qquad \mathrm{(B) \ } -8\qquad \mathrm{(C) \ } 0\qquad \mathrm{(D) \ } 8\qquad \mathrm{(E) \ } 20 </math>
 
<math> \mathrm{(A) \ } -16\qquad \mathrm{(B) \ } -8\qquad \mathrm{(C) \ } 0\qquad \mathrm{(D) \ } 8\qquad \mathrm{(E) \ } 20 </math>
  
==Solution==
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==Solution 1==
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>
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<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>
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Two [[polynomial]]s are equal only if their [[coefficient]]s are equal, so we must have
 
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>
 
<math>m^2 = 4, n^2 = 9</math>
 +
 
<math>m = \pm 2, n = \pm 3</math>
 
<math>m = \pm 2, n = \pm 3</math>
 +
 
<math>a + 8= 2mn = \pm 2\cdot 2\cdot 3 = \pm 12</math>
 
<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>a = 4</math> or <math>a = -20</math>.
  
 
So the desired sum is <math> (4)+(-20)=-16 \Longrightarrow \mathrm{(A)} </math>
 
So the desired sum is <math> (4)+(-20)=-16 \Longrightarrow \mathrm{(A)} </math>
 
==See Also==
 
*[[2005 AMC 10A Problems]]
 
  
*[[2005 AMC 10A Problems/Problem 9|Previous Problem]]
 
  
*[[2005 AMC 10A Problems/Problem 11|Next Problem]]
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Alternatively, note that whatever the two values of <math>a</math> are, they must lead to equations of the form <math>px^2 + qx + r =0</math> and <math>px^2 - qx + r = 0</math>.  So the two choices of <math>a</math> must make <math>a_1 + 8 = q</math> and <math>a_2 + 8 = -q</math> so <math>a_1 + a_2 + 16 = 0</math> and <math>a_1 + a_2 = -16\Longrightarrow \mathrm{(A)}</math>.
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==Solution 2==
 +
 
 +
Since this quadratic must have a double root, the discriminant of the quadratic formula for this quadratic must be 0. Therefore, we must have
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<cmath> (a+8)^2 - 4(4)(9) = 0 \implies a^2 + 16a - 80. </cmath> We can use the quadratic formula to solve for its roots (we can ignore the things in the radical sign as they will cancel out due to the <math>\pm</math> sign when added). So we must have
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<cmath> {-16 + \sqrt{\text{something}}}{2} + \frac{-16 - \sqrt{\text{something}}}{2}. </cmath>
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Therefore, we have <math>\implies \boxed{A}</math>.
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== Solution 3==
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There is only one positive value for k such that the quadratic equation would have only one solution.
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k-8 and -k-8 are the values of a.-8-8 is -16, so the answer is...<math>\implies \boxed{A}.</math>
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==See also==
 +
{{AMC10 box|year=2005|ab=A|num-b=9|num-a=11}}
 +
 
 +
{{MAA Notice}}

Latest revision as of 15:34, 8 December 2020

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 1

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 $a_1 + a_2 = -16\Longrightarrow \mathrm{(A)}$.

Solution 2

Since this quadratic must have a double root, the discriminant of the quadratic formula for this quadratic must be 0. Therefore, we must have \[(a+8)^2 - 4(4)(9) = 0 \implies a^2 + 16a - 80.\] We can use the quadratic formula to solve for its roots (we can ignore the things in the radical sign as they will cancel out due to the $\pm$ sign when added). So we must have \[{-16 + \sqrt{\text{something}}}{2} + \frac{-16 - \sqrt{\text{something}}}{2}.\] Therefore, we have $\implies \boxed{A}$.

Solution 3

There is only one positive value for k such that the quadratic equation would have only one solution. k-8 and -k-8 are the values of a.-8-8 is -16, so the answer is...$\implies \boxed{A}.$

See also

2005 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 9
Followed by
Problem 11
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 10 Problems and Solutions

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