Difference between revisions of "2021 Fall AMC 10A Problems/Problem 20"
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+ | A quadratic equation has one solution if and only if <math>\sqrt {b^2-4ax}</math> is <math>0</math>. Similarly, it is imaginary if and only if <math>\sqrt {b^2-4ax}</math> is less than one. 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 b and c. 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 {(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 b and c. 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 {(B) 6}</math> |
Revision as of 19:11, 22 November 2021
Problem
How many ordered pairs of positive integers exist where both
and
do not have distinct, real solutions?
Solution
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
Note that:
- If
then
from which
- If
then
from which
- If
then
from which
- If
then
from which
Together, there are
~MRENTHUSIASM
Solution 1(Oversimplified but risky)
A quadratic equation has one solution if and only if is
. Similarly, it is imaginary if and only if
is less than one. 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 b and c. 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.