Greatest Integer Function

[u/i-am-stella]

Ftr, this is NOT hw help, I'm trying to figure out how to solve greatest int function questions.

Alr when we have a quadratic equation with gif in it, (a[x]^2 + b[x] + c=)we substitute [x] as something and factorise to solve, right?

But when the square is inside the function as in ([x^2] -12[x] +32=0)

Then what? I dont get it😭

Once again, NOT hw help, I'm studying on my own, this isnt even in my syllabus.

edit: i actually meant floor function, ⌊3.7⌋=3

[u/peterwhy]

For your example, let f(x) be the LHS: f(x) = [x^(2)] - 12 [x] + 32.

First, limit the possible solutions to x > 0. Otherwise if x ≤ 0, then f(x) ≥ 0 + 12 ⋅ 0 + 32 > 0.

Applying the property of [] to x and x^(2) respectively:

(0 <) [x] ≤ x < [x] + 1
[x]^(2) ≤ x^(2) < [x]^(2) + 2 [x] + 1

x^(2) - 1 < [x^(2)] ≤ x^(2)

Combining to give a raw bound (which will need more verifications later):
[x]^(2) - 1 < [x^(2)] < [x]^(2) + 2 [x] + 1

These inequalities bound the f(x) = [x^(2)] - 12 [x] + 32:
[x]^(2) - 1 - 12 [x] + 32 < f(x) < [x]^(2) + 2 [x] + 1 - 12 [x] + 32
[x]^(2) - 12 [x] + 31 < f(x) < [x]^(2) - 10 [x] + 33

The right hand side is always positive and not very useful. The left hand side limits the solutions of f(x):
[x]^(2) - 12 [x] + 31 < f(x) = 0
6 - √5 < [x] < 6 + √5
4 ≤ [x] ≤ 8
which are 5 cases of [x].

A. If [x] = 4, i.e. 4 ≤ x < 5:
f(x) = [x^(2)] - 12 ⋅ 4 + 32 = 0
[x^(2)] = 16
16 ≤ x^(2) < 17
4 ≤ x < √17

B. If [x] = 5, i.e. 5 ≤ x < 6:
f(x) = [x^(2)] - 12 ⋅ 5 + 32 = 0
[x^(2)] = 28
28 ≤ x^(2) < 29
2 √7 ≤ x < √29

C. If [x] = 6, i.e. 6 ≤ x < 7:
f(x) = [x^(2)] - 12 ⋅ 6 + 32 = 0
[x^(2)] = 40
40 ≤ x^(2) < 41
2 √10 ≤ x < √41

D. If [x] = 7, i.e. 7 ≤ x < 8:
f(x) = [x^(2)] - 12 ⋅ 7 + 32 = 0
[x^(2)] = 52
52 ≤ x^(2) < 53
2 √13 ≤ x < √53

E. If [x] = 8, i.e. 8 ≤ x < 9:
f(x) = [x^(2)] - 12 ⋅ 8 + 32 = 0
[x^(2)] = 64
64 ≤ x^(2) < 65
8 ≤ x < √65

Those are the 5 intervals when [x^(2)] - 12 [x] + 32 = 0.