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

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==Problem==
 
==Problem==
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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> \textbf{(A) }-16\qquad\textbf{(B) }-8\qquad\textbf{(C) } 0\qquad\textbf{(D) }8\qquad\textbf{(E) }20 </math>
+
<math>
 +
\textbf{(A) } -16 \qquad\textbf{(B) } -8 \qquad\textbf{(C) } 0 \qquad\textbf{(D) } 8 \qquad\textbf{(E) } 20
 +
</math>
  
 
==Solution 1==
 
==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 <math>1</math> distinct root if and only if the left-hand side is a perfect square. So we require
  
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<math>4x^2 + ax + 8x + 9 = (mx + n)^2</math>
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<cmath>4x^2 + ax + 8x + 9 = (mx + n)^2 = m^2 x^2 + 2mnx + n^2.</cmath>
  
−
<math>4x^2 + ax + 8x + 9 = m^2x^2 + 2mnx + n^2</math>
+
Two polynomials are equal only if their coefficients are equal, so we must have <math>m^2 = 4</math> and <math>n^2 = 9</math>, which reduce to <math>m = \pm 2</math> and <math>n = \pm 3</math> respectively. By equating <math>x</math>-coefficients, it follows that <math>a + 8 = 2mn = \pm 2 \cdot 2\cdot 3 = \pm 12</math>, so either <math>a = 12-8 = 4</math> or <math>a = -12-8 = -20</math>.
  
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Two [[polynomial]]s are equal only if their [[coefficient]]s are equal, so we must have
+
Accordingly, the desired sum is <math>4+(-20) = \boxed{\textbf{(A) } -16}</math>.
  
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<math>m^2 = 4, n^2 = 9</math>
+
(Alternatively, we can observe that whatever the <math>2</math> 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>, corresponding to the perfect squares <math>\left(mx \pm n\right)^2 = 0</math>. So the <math>2</math> choices of <math>a</math>, say <math>a_1</math> and <math>a_2</math>, must satisfy <math>a_1 + 8 = q</math> and <math>a_2 + 8 = -q</math>, and adding these equations yields <math>a_1 + a_2 + 16 = 0 \iff a_1 + a_2 = \boxed{\textbf{(A) } -16}</math>.)
−
 
 
−
<math>m = \pm 2, n = \pm 3</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>.
 
−
 
 
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So the desired sum is <math> (4)+(-20)=\boxed{\textbf{(A)}-16} </math>
 
−
 
 
−
 
 
−
 
 
−
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==
 
==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
+
Since this quadratic must have a double root, its discriminant must be <math>0</math>. Therefore we require
−
<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>(a+8)^2 - 4 \cdot 4 \cdot  9 = 0 \iff a^2 + 16a - 80 = 0.</cmath> At this point, Vieta's formulae directly give the sum of the <math>2</math> possible values of <math>a</math> as <math>-\frac{16}{1} = \boxed{\textbf{(A) } -16}.</math>
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<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>
 
−
 
 
−
== 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>
 
  
 +
(Alternatively, we could use the quadratic formula to directly solve this equation for <math>a</math>; the precise value of the expression under the square root does not actually matter, as it will cancel out due to the <math>\pm</math> signs when added. This again gives the required sum as
 +
<cmath>\frac{-16 + \sqrt{\text{something}}}{2} + \frac{-16 - \sqrt{\text{something}}}{2} = \frac{-32}{2} = \boxed{\textbf{(A) } -16}.)</cmath>
 
==See also==
 
==See also==
 
{{AMC10 box|year=2005|ab=A|num-b=9|num-a=11}}
 
{{AMC10 box|year=2005|ab=A|num-b=9|num-a=11}}
  
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 17:24, 1 July 2025

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 $1$ distinct root if and only if the left-hand side is a perfect square. So we require

\[4x^2 + ax + 8x + 9 = (mx + n)^2 = m^2 x^2 + 2mnx + n^2.\]

Two polynomials are equal only if their coefficients are equal, so we must have $m^2 = 4$ and $n^2 = 9$, which reduce to $m = \pm 2$ and $n = \pm 3$ respectively. By equating $x$-coefficients, it follows that $a + 8 = 2mn = \pm 2 \cdot 2\cdot 3 = \pm 12$, so either $a = 12-8 = 4$ or $a = -12-8 = -20$.

Accordingly, the desired sum is $4+(-20) = \boxed{\textbf{(A) } -16}$.

(Alternatively, we can observe that whatever the $2$ values of $a$ are, they must lead to equations of the form $px^2 + qx + r = 0$ and $px^2 - qx + r = 0$, corresponding to the perfect squares $\left(mx \pm n\right)^2 = 0$. So the $2$ choices of $a$, say $a_1$ and $a_2$, must satisfy $a_1 + 8 = q$ and $a_2 + 8 = -q$, and adding these equations yields $a_1 + a_2 + 16 = 0 \iff a_1 + a_2 = \boxed{\textbf{(A) } -16}$.)

Solution 2

Since this quadratic must have a double root, its discriminant must be $0$. Therefore we require \[(a+8)^2 - 4 \cdot 4 \cdot 9 = 0 \iff a^2 + 16a - 80 = 0.\] At this point, Vieta's formulae directly give the sum of the $2$ possible values of $a$ as $-\frac{16}{1} = \boxed{\textbf{(A) } -16}.$

(Alternatively, we could use the quadratic formula to directly solve this equation for $a$; the precise value of the expression under the square root does not actually matter, as it will cancel out due to the $\pm$ signs when added. This again gives the required sum as \[\frac{-16 + \sqrt{\text{something}}}{2} + \frac{-16 - \sqrt{\text{something}}}{2} = \frac{-32}{2} = \boxed{\textbf{(A) } -16}.)\]

See also

2005 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 9 		Followed by
Problem 11
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All AMC 10 Problems and Solutions

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