Difference between revisions of "2002 AMC 12B Problems/Problem 6"
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+ | {{duplicate|[[2002 AMC 12B Problems|2002 AMC 12B #6]] and [[2002 AMC 10B Problems|2002 AMC 10B #10]]}} | ||
== Problem == | == 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 | 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 | ||
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\qquad\mathrm{(D)}\ (2,-1) | \qquad\mathrm{(D)}\ (2,-1) | ||
\qquad\mathrm{(E)}\ (4,4)</math> | \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) = | + | == Solution 1 == |
− | 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) \ | + | 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> | ||
+ | ==Solution 3 == | ||
+ | Since <math>a</math> and <math>b</math> are solutions to the given equation, we can write the two equations <math>a^2 + a^2 + b = 2a^2 + b = 0,</math> and <math>b^2 + ab + b = 0.</math> From the first equation, we get that <math>b = -2a^2,</math> and substituting this in our second equation, we get that <math>4a^4 - 2a^3 - 2a^2 = 0,</math> and solving this gives us the solutions <math>a = -\tfrac{1}{2},</math> <math>a = 0</math> and <math>a = 1.</math> We discard the first two solutions, as the first one doesnt show up in the answer choices and we are given that <math>a</math> is nonzero. The only valid solution is then <math>a = 1,</math> which gives us <math>b = -2,</math> and <math>(a, b) = \boxed{\text{(C) } (1, -2)}</math> | ||
+ | ==Solution 4 (Using the Answer Choices) == | ||
+ | Note that for roots <math>a</math> and <math>b</math>, <math>ab = b</math>. This implies that <math>a</math> is <math>1.0</math>, and there is only one answer choice with <math>1</math> in the position for <math>a</math>, hence, <math>(a,b) = \boxed{\mathrm{(C)}\ (1,-2)}</math> | ||
+ | |||
+ | == Video Solution == | ||
+ | https://youtu.be/5QdPQ3__a7I?t=498 | ||
+ | |||
+ | ~ pi_is_3.14 | ||
+ | |||
+ | https://www.youtube.com/watch?v=-3TRz6NTwCI ~David | ||
== See also == | == 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}} | {{AMC12 box|year=2002|ab=B|num-b=5|num-a=7}} | ||
[[Category:Introductory Algebra Problems]] | [[Category:Introductory Algebra Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 14:53, 17 August 2023
- The following problem is from both the 2002 AMC 12B #6 and 2002 AMC 10B #10, so both problems redirect to this page.
Contents
Problem
Suppose that and
are nonzero real numbers, and that the equation
has solutions
and
. Then the pair
is
Solution 1
Since , it follows by comparing coefficients that
and that
. Since
is nonzero,
, and
. Thus
.
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 coefficient, since the leading coefficient is 1; in other words,
and the product of the solutions is equal to the constant term (i.e,
). Since
is nonzero, it follows that
and therefore (from the first equation),
. Hence,
Solution 3
Since and
are solutions to the given equation, we can write the two equations
and
From the first equation, we get that
and substituting this in our second equation, we get that
and solving this gives us the solutions
and
We discard the first two solutions, as the first one doesnt show up in the answer choices and we are given that
is nonzero. The only valid solution is then
which gives us
and
Solution 4 (Using the Answer Choices)
Note that for roots and
,
. This implies that
is
, and there is only one answer choice with
in the position for
, hence,
Video Solution
https://youtu.be/5QdPQ3__a7I?t=498
~ pi_is_3.14
https://www.youtube.com/watch?v=-3TRz6NTwCI ~David
See also
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.