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

Revision as of 11:38, 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 $\frac{-16 + \sqrt{\text{something}}}{2} + \frac{-16 - \sqrt{\text{something}}}{2}.$ $\frac{-32}{2} = \boxed{\textbf{(A)}-16}$

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 $\boxed{\textbf{(A)} -16}$

@@ Line 35: / Line 35: @@
 == Solution 3==
-There is only one positive value for k such that the quadratic equation would have only one solution.
+There is only one positive value for <math>k</math> 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...<math>\implies \boxed{A}.</math>
+<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>
 ==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

Page

Toolbox

Search