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

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There are two values of <math>a</math> for which the equation <math> 4x^2 + ax + 8x + 9 = 0 </math> has only one solution for <math>x</math>. What is the sum of those values of <math>a</math>?
 
There are two values of <math>a</math> for which the equation <math> 4x^2 + ax + 8x + 9 = 0 </math> has only one solution for <math>x</math>. What is the sum of those values of <math>a</math>?
  
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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>
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<math> \textbf{(A) }-16\qquad\textbf{(B) }-8\qquad\textbf{(C) } 0\qquad\textbf{(D) }8\qquad\textbf{(E) }20 </math>
  
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==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
 
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>
 +
 
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
 +
 
<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>
 +
 
<math>a = 4</math> or <math>a = -20</math>.
 
<math>a = 4</math> or <math>a = -20</math>.
  
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So the desired sum is <math> (4)+(-20)=-16 \Longrightarrow \mathrm{(A)} </math>
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So the desired sum is <math> (4)+(-20)=\boxed{\textbf{(A)}-16} </math>
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==See Also==
+
 
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*[[2005 AMC 10A Problems]]
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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 =\boxed{\textbf{(A)}-16} </math>
 +
 
 +
==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
 +
<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 expression in the radical sign as it will cancel out due to the <math>\pm</math> sign when added). So we must have
 +
<cmath> \frac{-16 + \sqrt{\text{something}}}{2} + \frac{-16 - \sqrt{\text{something}}}{2}. </cmath>
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<math>\frac{-32}{2} = \boxed{\textbf{(A)}-16}</math>
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Note: you can also use Vieta's Formulae
 +
 
 +
== Solution 3==
 +
There is only one positive value for <math>k</math> such that the quadratic equation would have only one solution.
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<math>k-8</math> and <math>-k-8</math> are the values of <math>a</math>. <math>-8-8=-16</math>, so the answer is <math>\boxed{\textbf{(A)} -16}</math>
 +
 
 +
Another way of thinking of this is letting the two values of k be <math>a</math> and <math>a_0</math>. Since this is obviously a square of a binomial, <math>k+8 = m</math> or <math>-m</math> for some <math>m</math>. Thus, we can say that <math>a + 8 = m</math> and <math>a_0 + 8 = -m</math>. Combining these gives us <math>a_0 + a + 16 = 0</math>, so <math>a_0 + a = 16</math>. Our answer is <math>\boxed{\textbf{(A)} -16}</math>.
 +
 
 +
~Extremelysupercooldude (Second solution)
  
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*[[2005 AMC 10A Problems/Problem 9|Previous Problem]]
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==See also==
 +
{{AMC10 box|year=2005|ab=A|num-b=9|num-a=11}}
  
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*[[2005 AMC 10A Problems/Problem 11|Next Problem]]
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{{MAA Notice}}

Latest revision as of 19:33, 13 September 2024

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 $a$?

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

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 $(4)+(-20)=\boxed{\textbf{(A)}-16}$


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 $a_1 + a_2 =\boxed{\textbf{(A)}-16}$

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 expression in the radical sign as it will cancel out due to the $\pm$ sign when added). So we must have \[\frac{-16 + \sqrt{\text{something}}}{2} + \frac{-16 - \sqrt{\text{something}}}{2}.\] $\frac{-32}{2} = \boxed{\textbf{(A)}-16}$ Note: you can also use Vieta's Formulae

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 $a$. $-8-8=-16$, so the answer is $\boxed{\textbf{(A)} -16}$

Another way of thinking of this is letting the two values of k be $a$ and $a_0$. Since this is obviously a square of a binomial, $k+8 = m$ or $-m$ for some $m$. Thus, we can say that $a + 8 = m$ and $a_0 + 8 = -m$. Combining these gives us $a_0 + a + 16 = 0$, so $a_0 + a = 16$. Our answer is $\boxed{\textbf{(A)} -16}$.

~Extremelysupercooldude (Second solution)

See also

2005 AMC 10A (Problems • Answer Key • Resources)
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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