Difference between revisions of "2021 Fall AMC 10A Problems/Problem 20"
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==Solution 1(Oversimplified but risky)== | ==Solution 1(Oversimplified but risky)== | ||
| − | 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 | + | 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 ![\[b^4\leq 16c^2\leq 64b.\]](http://latex.artofproblemsolving.com/c/f/8/cf8fa6bf7c791041cdad74405e8789551ecc94c1.png)
Note that:
- If
then 
from which 
then 
from which 
- If
then 
from which
then 
from which - If
then 
from which 
then 
from which 
- If
then 
from which 
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. 
