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Equation - Quadratic Equation

For COMPETITION
Number of Total Problems: 11.
FOR PRINT ::: (Book)

Problem Num : 11

From : AMC10

Type:

Section:Equation

Theme:
Adjustment# : 0

Difficulty: 1

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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 $a$ ?
$mathrm{(A) } -16qquad mathrm{(B) } -8qquad mathrm{(C) } 0qquad mathrm{(D) } 8qquad mathrm{(E) } 20$
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Category Quadratic Equation

Analysis

Solution/Answer
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
$m^2 = 4, n^2 = 9$
$m = pm 2, n = pm 3$
$a + 8= 2mn = pm 2cdot 2cdot 3 = pm 12$
$a = 4$ 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 $a_1 + a_2 = -16Longrightarrow mathrm{(A)}$ .

Alternate Solution

Since this quadratic must have a double root, the discriminant of the quadratic formula for this quadratic must be 0. Therefore, we must have $(a+8)^2 - 4(4)(9) = 0 implies a^2 + 16a - 144.$ 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 $frac{-16 + sqrt{ ext{something}}}{2} + frac{-16 - sqrt{ ext{something}}}{2}.$ Therefore, we have $(-16)(2)/2 = -16 implies oxed{A}.$

Answer:

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