2007 AMC 12A Problems/Problem 21

Problem

The sum of the zeros, the product of the zeros, and the sum of the coefficients of the function $\displaystyle f(x)=ax^{2}+bx+c$ are equal. Their common value must also be which of the following?

$\textrm{(A)}\ \textrm{the\ coefficient\ of\ }x^{2}~~~ \textrm{(B)}\ \textrm{the\ coefficient\ of\ }x$ $\textrm{(C)}\ \textrm{the\ y-intercept\ of\ the\ graph\ of\ }y=f(x)$ $\textrm{(D)}\ \textrm{one\ of\ the\ x-intercepts\ of\ the\ graph\ of\ }y=f(x)$ $\textrm{(E)}\ \textrm{the\ mean\ of\ the\ x-intercepts\ of\ the\ graph\ of\ }y=f(x)$

Solution

By Vieta's formulas, the sum of the roots of a quadratic equation is $\frac {-b}a$ , the product of the zeros is $\frac ca$ , and the sum of the coefficients is $a + b + c$ . Setting equal the first two tells us that $\frac {-b}{a} = \frac ca \Rightarrow b = -c$ . Thus, $a + b + c = a + b - b = a$ , so the common value is also equal to the coefficient of $x^2 \Longrightarrow \textrm{A}$ .

To disprove the others, note that:

$\mathrm{B}$ : then $b = \frac {-b}a$ , which forces $a = -1$ (not an identity).
$\mathrm{C}$ : the y-intercept is $c$ , so $c = \frac ca$ which forces $a = 1$ .
$\mathrm{D}$ : an x-intercept of the graph is a root of the polynomial, but this excludes the other root.
$\mathrm{E}$ : the mean of the x-intercepts will be the sum of the roots of the quadratic divided by 2.

2007 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 20	Followed by Problem 22
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
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2007 AMC 12A Problems/Problem 21

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