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2005 AMC 10A Problems/Problem 10

Revision as of 19:05, 17 September 2019 by Boddapatis (talk | contribs) (→‎Solution 3)

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 things in the radical sign as they will cancel out due to the $\pm$ sign when added). So we must have \[{-16 + \sqrt{\text{something}}}{2} + \frac{-16 - \sqrt{\text{something}}}{2}.\] 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 $\implies \boxed{A}.$

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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