Difference between revisions of "2023 AMC 10B Problems/Problem 22"
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− | == Solution == | + | == Solution (Quick) == |
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+ | 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> | ||
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+ | So we got 4 in total <math>(\dfrac{2}{3},1,2,\dfrac{11}{3}).</math> | ||
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+ | ~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