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Difference between revisions of "2002 AMC 12B Problems/Problem 6"

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== Solution ==
 
== Solution ==
 
Since <math>(x-a)(x-b) = x^2 - (a+b)x + ab = x^2 + ax + b = 0</math>, it follows by comparing [[coefficient]]s that <math>-a - b = a</math> and that <math>ab = b</math>. Since <math>b</math> is nonzero, <math>a = 1</math>, and <math>-1 - b = 1 \Longrightarrow b = -2</math>. Thus <math>(a,b) = (1,-2) \Rightarrow \mathrm{(C)}</math>.
 
Since <math>(x-a)(x-b) = x^2 - (a+b)x + ab = x^2 + ax + b = 0</math>, it follows by comparing [[coefficient]]s that <math>-a - b = a</math> and that <math>ab = b</math>. Since <math>b</math> is nonzero, <math>a = 1</math>, and <math>-1 - b = 1 \Longrightarrow b = -2</math>. Thus <math>(a,b) = (1,-2) \Rightarrow \mathrm{(C)}</math>.
 +
Another method is to use [[Vieta's formulas]]. The sum of the solutions to this polynomial is equal to the opposite of the <math>x</math> coefficient, since the leading coefficient is 1; in other words, <math>a + b = -a</math> and the product of the solutions is equal to the constant term (i.e, <math>a*b = b</math>). Since <math>b</math> is nonzero, it follows that <math>a = 1</math> and therefore (from the first equation), <math>b = -2a = -2</math>. Hence, <math>(a,b) = (1,-2) \Rightarrow \mathrm{ (C)}</math>
  
 
== See also ==
 
== See also ==

Revision as of 20:40, 29 June 2008

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

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 $(a,b) = (1,-2) \Rightarrow \mathrm{(C)}$. 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, $(a,b) = (1,-2) \Rightarrow \mathrm{ (C)}$

See also

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
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