2023 AMC 10B Problems/Problem 22: Difference between revisions
Technodoggo (talk | contribs) |
Technodoggo (talk | contribs) |
||
| Line 1: | Line 1: | ||
== Solution == | == Solution (Quick) == | ||
A quadric equations can have up to 2 real solutions. With the <math>\lfloor{x}\rfloor</math>, it could also help generate another pair. We have to verify that the solutions are real and distinct. | |||
First, we get the trivial solution by ignoring the floor. | First, we get the trivial solution by ignoring the floor. | ||
<math>(x-2)(x-1) = 0</math>, we get <math>(2,1)</math> as solutions. | <math>(x-2)(x-1) = 0</math>, we get <math>(2,1)</math> as our first pair of solutions. | ||
Up to this point, we can rule out A,E. | |||
Next, we see that <math>\lfloor{x}\rfloor^2-3x=0.</math> This implies that <math>-3x</math> must be an integer. | Next, we see that <math>\lfloor{x}\rfloor^2-3x=0.</math> This implies that <math>-3x</math> must be an integer. | ||
We can guess and check <math>x</math> as <math>\dfrac{k}{3}</math> which yields <math>(\dfrac{2}{3},\dfrac{11}{3}).</math> | We can guess and check <math>x</math> as <math>\dfrac{k}{3}</math> which yields <math>(\dfrac{2}{3},\dfrac{11}{3}).</math> | ||
So we got 4 in total <math>(\dfrac{2}{3},1,2,\dfrac{11}{3}).</math> | |||
~Technodoggo | |||
Revision as of 14:02, 15 November 2023
Solution (Quick)
A quadric equations can have up to 2 real solutions. With the 
, it could also help generate another pair. We have to verify that the solutions are real and distinct.
First, we get the trivial solution by ignoring the floor.
, we get 
as our first pair of solutions.
Up to this point, we can rule out A,E.
Next, we see that 
This implies that 
must be an integer.
We can guess and check 
as 
which yields 
So we got 4 in total 
~Technodoggo
