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 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 $1.0$ , and there is only one answer choice with $1$ in the position for $a$ , hence, $(a,b) = \boxed{\mathrm{(C)}\ (1,-2)}$

Video Solution

https://youtu.be/5QdPQ3__a7I?t=498

~ pi_is_3.14

https://www.youtube.com/watch?v=-3TRz6NTwCI ~David

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.

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