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

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

Solutions

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

Since $a$ and $b$ are solutions to the given equation, we can write the two equations $a^2 + a^2 + b = 2a^2 + b = 0,$ and $b^2 + ab + b = 0.$ From the first equation, we get that $b = -2a^2,$ and substituting this in our second equation, we get that $4a^4 - 2a^3 - 2a^2 = 0,$ and solving this gives us the solutions $a = -\tfrac{1}{2},$ $a = 0$ and $a = 1.$ We discard the first two solutions, as the first one doesnt show up in the answer choices and we are given that $a$ is nonzero. The only valid solution is then $a = 1,$ which gives us $b = -2,$ and $(a, b) = \boxed{\text{(C) } (1, -2)}$

Solution 4 (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, $(a,b) = \boxed{\mathrm{(C)}\ (1,-2)}$

See also

2002 AMC 10B (ProblemsAnswer KeyResources)
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 (ProblemsAnswer KeyResources)
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

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