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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>
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== Solution ==
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−
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>.
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== 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) \Rightarrow \mathrm{ (C)}</math>
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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>.
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==Solution 2 ==
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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>
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==Solution 3 ==
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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>
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==Solution 4 (Using the Answer Choices) ==
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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>
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==Solution 5(Answer Choices) ==
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We can use Vieta's formulas to solve this. We know that <math>a + b = -a</math>, and <math>ab = b.</math> We can test the answer choices, so that we find that <math>\boxed{\mathrm{(C)}\ (1,-2)}</math> works.
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 +
-monkey_land
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== Video Solution ==
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https://youtu.be/5QdPQ3__a7I?t=498
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 +
~ pi_is_3.14
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https://www.youtube.com/watch?v=-3TRz6NTwCI  ~David
  
 
== See also ==
 
== See also ==
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{{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]]
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{{MAA Notice}}

Latest revision as of 12:14, 3 November 2024

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

Solution 5(Answer Choices)

We can use Vieta's formulas to solve this. We know that $a + b = -a$, and $ab = b.$ We can test the answer choices, so that we find that $\boxed{\mathrm{(C)}\ (1,-2)}$ works.

-monkey_land

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

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