Difference between revisions of "2021 Fall AMC 10A Problems/Problem 20"
MRENTHUSIASM (talk | contribs) |
MRENTHUSIASM (talk | contribs) |
||
Line 6: | Line 6: | ||
== Solution == | == Solution == | ||
− | A quadratic equation | + | A quadratic equation does not have real solutions if and only if the discriminant is nonpositive. We conclude that: |
+ | <ol style="margin-left: 1.5em;"> | ||
+ | <li>Since <math>x^2+bx+c=0</math> does not have real solutions, we have <math>b^2\leq 4c.</math></li><p> | ||
+ | <li>Since <math>x^2+cx+b=0</math> does not have real solutions, we have <math>c^2\leq 4b.</math></li><p> | ||
+ | </ol> | ||
+ | Squaring the first inequality, we get <math>b^4\leq 16c^2.</math> Multiplying the second inequality by <math>16,</math> we get <math>16c^2\leq 64b.</math> Combining these results, we get <cmath>b^4\leq 16c^2\leq 64b.</cmath> | ||
+ | |||
+ | ==See Also== | ||
+ | {{AMC10 box|year=2021 Fall|ab=A|num-b=19|num-a=21}} | ||
+ | {{MAA Notice}} |
Revision as of 19:08, 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
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.