x
2 + x – 12 = 0
⇒ x
2 + 4x – 3x – 12 = 0
⇒ x(x + 4) – 3(x + 4) = 0
⇒ (x – 3)(x + 4) = 0
⇒ x – 3 = 0 or x + 4 = 0
⇒ x = 3 or x = -4
therefore, the solution set of the given equation can be written in roaster form as {3, -4}
So, J = {3, -4}
Q. 3 J. Write the following sets iroster from:
K = {x : ϵ N, x is a multiple of 5 and x2 < 400}.
Answer : Multiple of 5 are 5, 10, 15, 20, 25, 30, …
So, 52 = 25
102 = 100
152 = 225