x2−7x+10≤0
Solve for x2−7x+10=0
We have, (x - 5)(x - 2) ≤ 0.
Conditions:
Case 1: (x - 5) ≤ 0, (x - 2) ≥ 0.
implies x ≤ 5; x ≥ 2.
2 ≤ x ≤ 5.
Choosing x = 3,
32 - 7(3) + 10 = 9 - 21 + 10
= -2 ≤ 0.
∴ 2 ≤ x ≤ 5.