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

Revision as of 18:00, 28 July 2011

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

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

@@ Line 1: / Line 1: @@
+{{duplicate|[[2002 AMC 12B Problems|2002 AMC 12B #6]] and [[2002 AMC 10B Problems|2002 AMC 10B #10]]}}
 == Problem ==
 Suppose that <math>a</math> and <math>b</math> are nonzero real numbers, and that the [[equation]] <math>x^2 + ax + b = 0</math> has solutions <math>a</math> and <math>b</math>. Then the pair <math>(a,b)</math> is
@@ Line 8: / Line 9: @@
 \qquad\mathrm{(E)}\ (4,4)</math>
 == 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>.
-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>
+=== Solution 1 ===
+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) = \boxed{\mathrm{(C)}\ (1,-2)}</math>.
+===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 <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) = \boxed{\mathrm{(C)}\ (1,-2)}</math>
 == See also ==
+{{AMC10 box|year=2002|ab=B|num-b=9|num-a=11}}
 {{AMC12 box|year=2002|ab=B|num-b=5|num-a=7}}
 [[Category:Introductory Algebra Problems]]

Page

Toolbox

Search