x2−7x+10≤0x^2 - 7x + 10 \leq 0x2−7x+10≤0
Solve for x2−7x+10=0x^2 - 7x + 10 = 0x2−7x+10=0
We have, (x - 5)(x - 2) ≤\leq≤ 0.
Conditions:
Case 1: (x - 5) ≤\leq≤ 0, (x - 2) ≥\geq≥ 0.
⟹ \implies⟹ x ≤\leq≤ 5; x ≥\geq≥ 2.
2 ≤\leq≤ x ≤\leq≤ 5.
Choosing x = 3,
32^22 - 7(3) + 10 = 9 - 21 + 10
= -2 ≤\leq≤ 0.
∴\therefore∴ 2 ≤\leq≤ x ≤\leq≤ 5.
