2002 AMC 12B Problems/Problem 6

The following problem is from both the 2002 AMC 12B #6 and 2002 AMC 10B #10, so both problems redirect to this page.

Problem

Suppose that $a$ and $b$ are nonzero real numbers, and that the equation $x^2 + ax + b = 0$ has solutions $a$ and $b$ . Then the pair $(a,b)$ is

$\mathrm{(A)}\ (-2,1) \qquad\mathrm{(B)}\ (-1,2) \qquad\mathrm{(C)}\ (1,-2) \qquad\mathrm{(D)}\ (2,-1) \qquad\mathrm{(E)}\ (4,4)$

Solution

Solution 1

Since $(x-a)(x-b) = x^2 - (a+b)x + ab = x^2 + ax + b = 0$ , it follows by comparing coefficients that $-a - b = a$ and that $ab = b$ . Since $b$ is nonzero, $a = 1$ , and $-1 - b = 1 \Longrightarrow b = -2$ . Thus $(a,b) = \boxed{\mathrm{(C)}\ (1,-2)}$ .

Solution 2

Another method is to use Vieta's formulas. The sum of the solutions to this polynomial is equal to the opposite of the $x$ coefficient, since the leading coefficient is 1; in other words, $a + b = -a$ and the product of the solutions is equal to the constant term (i.e, $a*b = b$ ). Since $b$ is nonzero, it follows that $a = 1$ and therefore (from the first equation), $b = -2a = -2$ . Hence, $(a,b) = \boxed{\mathrm{(C)}\ (1,-2)}$

Solution 3 (Using the Answer Choices)

Note that for roots $a$ and $b$ , $ab = b$ . This implies that $a$ is $1$ , and there is only one answer choice with $1$ in the position for $a$ , hence, $(a,b) = \boxed{\mathrm{(C)}\ (1,-2)}$

2002 AMC 10B (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

2002 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 5	Followed by Problem 7
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 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

Page

Toolbox

Search