Question

Find the number of real values of x for which |x – 3| + (x – 3)^2 + √(x-3) + |x + 3| = 0.​

Answer 1

Answer:

[−1,2)∪[3,∞)

Solution :

Since, y=(x+1)(x−3)(x−2)−−−−−−−−−−−−√ takes all real values only

when (x+1)(x−3)(x−2)≥0

⇒−1≤x<2orx≥3

∴x∈[−1,2)∪[3,∞).

Explanation: