10) A = {x ∈ P | x ≤ 10} B = {x | x² + 3x - 40 = 0}
Answer (1)
To solve the question, we have two sets defined:
Set A is given by A = { x ∈ P ∣ x ≤ 10 } , where P represents some initial set, and we are considering the subset of P whose elements satisfy x ≤ 10 .
Set B is defined by the equation x 2 + 3 x − 40 = 0 . To find the elements of B , we need to solve this quadratic equation. We can use the quadratic formula:
x = 2 a − b ± b 2 − 4 a c
Here, a = 1 , b = 3 , and c = − 40 . Plugging these values into the formula, we get:
x = 2 × 1 − 3 ± 3 2 − 4 × 1 × ( − 40 )
Simplifying further,
x = 2 − 3 ± 9 + 160 = 2 − 3 ± 169
Since 169 = 13 , we find the roots as:
x 1 = 2 − 3 + 13 = 5 and x 2 = 2 − 3 − 13 = − 8
Therefore, set B is B = { 5 , − 8 } .
If we consider both sets, Set A takes values from the set P such that the elements are less than or equal to 10. Set B includes the solutions x = 5 and x = − 8 .
In conclusion, the values in Set B can be categorized within the specified range if they are part of the set P . We can't fully define Set A without knowing P , but any value that satisfies x ≤ 10 from B will also be in A if P refers to all real numbers.
