1) Find all x ∈ (All Real Numbers) such that x^2 - x + 6 = 0:
We can use the quadratic formula to solve for x:
x = (-b ± √(b^2 - 4ac)) / 2a
In this case, a = 1, b = -1, and c = 6, so we have:
x = (1 ± √(1 - 4(1)(6))) / 2(1)
x = (1 ± √(-23)) / 2
Since the square root of a negative number is not a real number, there are no real solutions to this equation. Therefore, the set of solutions is { } or the empty set.
2) Find all x ∈ (All Real Numbers) such that (x-3)²(2x+(√3))(x+5) = 0:
We can use the zero product property to find the solutions to this equation. This property states that if the product of two or more factors is equal to zero, then at least one of the factors must be equal to zero. Therefore, we need to solve the following three equations:
x - 3 = 0
2x + (√3) = 0
x + 5 = 0
Solving each of these equations, we get:
x = 3
x = -(√3) / 2
x = -5
Therefore, the set of solutions is {3, -(√3) / 2, -5}.
3) Find all even integers that are divisible by 5:
An integer is even if it is divisible by 2. Therefore, we need to find the integers that are divisible by both 2 and 5. This means we are looking for multiples of 10, which are even integers that end in 0. Therefore, the set of solutions is {10n : n ∈ (All Integers)}.