if [x] denotes the integral part of x, then what is the solution set of the equation [x]^2 -5[x]+6=0
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
To find the solution set of the equation [x]^2 - 5[x] + 6 = 0, we will use the concept of the floor function.
Let's start by considering the quadratic equation using the floor function:
[x]^2 - 5[x] + 6 = 0
Now, let's solve this equation by considering two cases:
Case 1: [x] is a non-negative integer
If [x] is a non-negative integer, then we can substitute [x] with a non-negative integer value.
Let's consider the possible values for [x]:
For [x] = 0: (0)^2 - 5(0) + 6 = 6 ≠ 0 (not a solution)
For [x] = 1: (1)^2 - 5(1) + 6 = 2 ≠ 0 (not a solution)
For [x] = 2: (2)^2 - 5(2) + 6 = 0 (a valid solution)
For [x] > 2, the equation becomes negative, so we don't need to consider further.
Therefore, in this case, the value of x = 2 is a solution.
Case 2: [x] is a negative integer
If [x] is a negative integer, then we can substitute [x] with a negative integer value.
Let's consider the possible values for [x]:
For [x] = -1: (-1)^2 - 5(-1) + 6 = 12 ≠ 0 (not a solution)
For [x] = -2: (-2)^2 - 5(-2) + 6 = 16 ≠ 0 (not a solution)
For [x] = -3: (-3)^2 - 5(-3) + 6 = 0 (a valid solution)
For [x] < -3, the equation becomes positive, so we don't need to consider further.
Therefore, in this case, the value of x = -3 is a solution.
Combining both cases, the solution set of the equation [x]^2 - 5[x] + 6 = 0 is {x = -3, x = 2}.