Difference between revisions of "2021 Fall AMC 10A Problems/Problem 20"
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==Solution 2 (Oversimplified but Risky)== | ==Solution 2 (Oversimplified but Risky)== | ||
− | A quadratic equation <math>Ax^2+Bx+C=0</math> has one solution if and only if <math>\sqrt{B^2-4AC}=0.</math> Similarly, it | + | A quadratic equation <math>Ax^2+Bx+C=0</math> has one real solution if and only if <math>\sqrt{B^2-4AC}=0.</math> Similarly, it has imaginary solutions if and only if <math>\sqrt{B^2-4AC}<0.</math> We proceed as following: |
We want both <math>x^2+bx+c</math> to be <math>1</math> value or imaginary and <math>x^2+cx+b</math> to be <math>1</math> value or imaginary. <math>x^2+4x+4</math> is one such case since <math>\sqrt {b^2-4ac}</math> is <math>0.</math> Also, <math>x^2+3x+3, x^2+2x+2, x^2+x+1</math> are always imaginary for both <math>b</math> and <math>c.</math> We also have <math>x^2+x+2</math> along with <math>x^2+2x+1</math> since the latter has one solution, while the first one is imaginary. Therefore, we have <math>6</math> total ordered pairs of integers, which is <math>\boxed{\textbf{(B) } 6}.</math> | We want both <math>x^2+bx+c</math> to be <math>1</math> value or imaginary and <math>x^2+cx+b</math> to be <math>1</math> value or imaginary. <math>x^2+4x+4</math> is one such case since <math>\sqrt {b^2-4ac}</math> is <math>0.</math> Also, <math>x^2+3x+3, x^2+2x+2, x^2+x+1</math> are always imaginary for both <math>b</math> and <math>c.</math> We also have <math>x^2+x+2</math> along with <math>x^2+2x+1</math> since the latter has one solution, while the first one is imaginary. Therefore, we have <math>6</math> total ordered pairs of integers, which is <math>\boxed{\textbf{(B) } 6}.</math> |
Revision as of 19:19, 22 November 2021
Contents
[hide]Problem
How many ordered pairs of positive integers exist where both
and
do not have distinct, real solutions?
Solution 1 (Casework)
A quadratic equation does not have real solutions if and only if the discriminant is nonpositive. We conclude that:
- Since
does not have real solutions, we have
- Since
does not have real solutions, we have
Squaring the first inequality, we get Multiplying the second inequality by
we get
Combining these results, we get
We apply casework to the value of
- If
then
from which
- If
then
from which
- If
then
from which
- If
then
from which
Together, there are ordered pairs
namely
and
~MRENTHUSIASM
Solution 2 (Oversimplified but Risky)
A quadratic equation has one real solution if and only if
Similarly, it has imaginary solutions if and only if
We proceed as following:
We want both to be
value or imaginary and
to be
value or imaginary.
is one such case since
is
Also,
are always imaginary for both
and
We also have
along with
since the latter has one solution, while the first one is imaginary. Therefore, we have
total ordered pairs of integers, which is
~Arcticturn
See Also
2021 Fall AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 19 |
Followed by Problem 21 | |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.