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

Revision as of 11:32, 13 December 2021

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 $\frac{-16 + \sqrt{\text{something}}}{2} + \frac{-16 - \sqrt{\text{something}}}{2}.$ $-32$ / $2$ = $-16$

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... $\implies \boxed{A}.$

@@ Line 21: / Line 21: @@
 <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)=\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 = -16\Longrightarrow \mathrm{(A)}</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==

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