Question

if [x] denotes the integral part of x, then what is the solution set of the equation [x]^2 -5[x]+6=0

Answer 1

x^2 - 5x + 6 = 0 where x = 2 or 3

for √2<x<2,
[x]^2 - 5[x] + 6 = 2-7+6 = 1 > 0
similarly for other smaller values of x

for 2<=x<√5,
[x]^2 - 5[x] + 6 = 4-10+6 = 0

√5<=x<√6, √6<x<√7, √7<x<√8, √8<x<3
[x]^2 - 5[x] + 6 > 0

for 3 <= x < √10,
[x]^2 - 5[x] + 6 = 9 - 14 + 6 = 0

for x>3, f(x) > 0

so, the solution set is

2<=x<√5 or 3<=x<√10