Question

The greatest integer function,[x] , is equal to the largest integer that is smaller than or equal to x. In the domain [0,100], how many solutions are there to x^2 - [x] x = 88.25?

Answer 1

Similar problem here:
http://openstudy.com/study#/updates/53081551e4b00fb1e23842b8

Answer 2

Basically, recall that: \[
0\leq x-\lfloor x \rfloor <1
\]

Answer 3

\[
0\leq \frac{88.25}{x} < 1
\]

Answer 4

This means we at least can realize:\[
88.25<x\leq 100
\]It's not feasible otherwise.

Answer 5

If we partition the interval \((88.25,100]\) based on \(\lfloor x \rfloor \), we can find a single solution in each partion.

Answer 6

Except maybe \(\lfloor x\rfloor = 100\) and \(\lfloor x\rfloor =88\). Those we'd have to double check.

Answer 7

11.75 is your answer, idk how but believe in the bat.

Answer 8

How can you have 0.75 of a solution? lol

Answer 9

12

Answer 10

okay hold on a second though.

Answer 11

I agree that \(\lfloor x \rfloor =89, 90, 91, \dots 99\) will have solutions for sure.

Answer 12

So we have 11 solutions. But then we need to figure out if there is a solution in \(\lfloor x\rfloor = 88\).

Answer 13

\[
x(x-88) = 88.25
\]

Answer 14

Hmm, that gives \(x = 88.9917\).... so I agree there is a 12th solution.

Answer 15

:p

Answer 16

okay now I'm wondering...

Answer 17

If we let \(0\leq b < 1\) and have \(n\)...
So instead of \(88.25\) we'd say \(b=0.25\), \(n=88\), \(88.25 = n+b\)

Answer 18

\[
x(x-n) = n+b
\] will this always have a solution?

Answer 19

That is, a real solution \(x_0\) that we know is \(n+b<x_0<n+1\)

Answer 20

let x=n+d, where n is an integer and 0<d<1
then x^2-x[x] = x(x-[x]) = (n+d)(n+d-n) = (n+d)d = 88.25
which will have solutions from n=88 to n=100
thus the number of solutions is 13

Answer 21

for n=88, we have
(88+d)d = 88.25
88d+d^2 = 88.25
let f(d)=88d+d^2
f(0)=0
f(1)=89
since f is continuous, for some value d between 0 and 1, f(d)=88.25

Answer 22

Well, \(d\) can't be under \(0.25\)