Solving a Quadratic Floor Equation

This video is about a floor equation
If you need to post a picture of your solution or idea:
#ChallengingMathProblems #FloorValueEquations

EXPLORE:

[neuralwarp]

Conjecture:
floor(x**a) = floor(x)**a
for values of x, a where
0 <= x**a < floor(x**a +1)
Needs refining

[sil]

Interesting ceiling/floor problems. Btw sqrt(2) at the end can be avoided, we have (n-2)^2<2, but only perfect squares less than 2 are 0^2 and 1^2, so we need to solve (n-2)^2=0 and (n-2)^2=1 (giving n=1, 2, 3)

[renatsamur]

Great video! I had another method where I said floor(x) =m therefore floor(x^2) = m^2 +n for non negative integers m, n. It follows n =0 or 1 and we can solve for x in each case. It was great to see another way of thinking from you! :)

[srijanbhowmick]

Very good and satisfying solution as well as the problem 😁

[roberttelarket]

The best way to solve this is on the floor not on the roof!!

[tgx]

We can also use floor (x^2)=x^2-{x^2}; floor(x)=x-{x}. We have then (x^2-4x+3)+(4{x}-{x^2})=0. But (4{x}-{x^2} )is defined on interval <0;1), there is minimum zero, supremum Is there 3, but function x^2-4x+3 +(4{x}-{x^2}) has roots for x in maximum interval <1;3>, So we can take cases a) floor (x)=1, then floor(x^2)=1. b)floor(x)=2, then floor(x^2)=5. c) flór(x)=3, then floor(x^2)=9. Then we have the same solution of this problem.

[saranyadas]

Amazing, just loved it, ❤️☺️☺️☺️☺️, thank you very much

[ben_adel]

i wish the question was (floorx)^2-4floorx+3 so we would factorize it like (floorx-3)(floorx-1)=0 that would have made the answer x (0, 1), (3, 4) but this intersting aswell and harder also

[damiennortier]

When you have found that x is between n and n + 1 and x^2 is between n^2 and (n+1)^2, you just have to use substitutation : floor value of x is u so the floor value of x^2 is u^2. Now, you just have to use discriminante : delta = - 4^2 - 4*3 = 4 so u1 = 1 and u2 = 3. Of course, x needs to be (or not to be 😂) between 1 and 2 or between 3 and 4. No?

[mxsjncv]

Thank you for showing that. If I want to explain it to a 9th grade good student, I will draw a nice y=x^2+3 graph, and y=4x graph. Show that they intersect at 1 and 3. Make lines parellal to x axis, length 1, from (0, 0), (1, 4), (2, 8), (3, 12), (4, 16), (5, 20). Lets start with the line from (1, 4). The parabulla changes to a line from x=1 to x=2^0.5 . So these lines are "together". Next two lines that are "together" are from x=5^0.5 to 6^0.5 ... you got the idea. At x=15^0.5 and on, they do not overlap any more.

[luiscrispinvargas]

Excelente video, me salió a la primera 🥹, un saludo desde Perú.

[leecherlarry]

i am upvoting every time i post a computer solution, like now:
*𝚁𝚎𝚍𝚞𝚌𝚎[𝙵𝚕𝚘𝚘𝚛[𝚡^𝟸] - 𝟺 𝙵𝚕𝚘𝚘𝚛[𝚡] + 𝟹 == 0, 𝚡, 𝚁𝚎𝚊𝚕𝚜]*

this way i know when my post was deleted when i revisit the video thanks! :P
btw please doht feel annoyed when i post computer solution. i actually post the working input only. it is interesting to see how we can formulate the problem in code. i doht post the output.
hope you doht mind!

[yqsixzi]

Thank you for your beautiful Floor Eqn

[tamarpeer]

we know frac(x) can be up to 1, so x^2 can be up to (round(x)+1)^2, and at least round(x)^2

(round(x)-3)(round(x)-1)<=0, round x is between 1 and 3
Second side doesn’t give much information
Let’s check each case
floor(x)=1, floor(x^2)-4+3=0, floor(x^2)=1, 1<=x<sqrt(2)
floor(x)=2, floor(x^2)-8+3=0
floor(x^2)=5
sqrt(5)<=x<sqrt(6)
floor(x)=3, floor(x^2)-12+3=0, floor(x^2)=9, 3<=x<sqrt(10)
Those are all the options

[hadibendj]

Marvelous solution, thank you very much sir.

[wdmwgnc]

Thank you! Please shoot a video about gcd properties

[pranavamali]

Square is always positive so square can also be taken outside mod just solve the quadratic u will get 4 solutions ±3, ±1

[riccardofroz]

Wow this one was tougher than expected actually:
\x^2/-4\x/+3=0

A: x=n+f with n integer and f in R, 0>=f<1
\x^2/=n^2+k with k integer 0<=k<=2n

n^2+k-4n+3=0
n^2-4n+3+k=0

n=(4+-sqrt(16-12-4k))/2
n=2+-sqrt(1-k)

1-k>=0
1>=k
0<=k<=1, k=1 or k=0

Expand \x^2/=n^2+k using A

\n^2+2nf+f^2/=n^2+k
n^2+\2nf+f^2/=n^2+k
\2nf+f^2/=k

k=0
n=1, then 0<=2f+f^2<1,
f<=-2, f>=0, -1-sqrt(2)<f<-1+sqrt(2) --> 0<=f<-1+sqrt(2)
1<=x<sqrt(2)
n=3, then 0<=6f+f^2<1,
f<=-6, f>=0, -3-sqrt(10)<f<-3+sqrt(10) --> 0<=f<-3+sqrt(10)
3<=x<sqrt(10)
k=1
n=2, then 1>=4f+f^2<2,
f<=-2-sqrt(5), f>=-2+sqrt(5), -2-sqrt(6)<f<-2+sqrt(6) --> -2+sqrt(5)<=f<-2+sqrt(6)
sqrt(5)<=x<sqrt(6)

[samuelringmuhuboh]

Thank you. Very good video.

Please can you look at this one

|X²|- |X|²=100,

[jimmykitty]

Yeap! I solved the problem 💖☺💖
Thanks a Zettabyte, Boss!! 🌿🌿🌿