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

Revision as of 12:14, 31 May 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 $a$ ?

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

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 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 32: / Line 32: @@
 <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>
--32/2=16
+<math>-32</math> / <math>2</math> = <math>16</math>
 Therefore, we have <math>\implies \boxed{A}</math>.

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