-8,-6,-4
-10,-6,0
these are the possible test points for your three intervals
The given inequality is, x² + 12 x+35≥0
→ x²+7 x +5 x+5×7≥0
→ x (x+7) +5(x+7)≥0
→ (x+5)(x+7)≥0
→ x+5=0 ∧ x+7=0 gives x= -7 ∧ x= -5
Now drawing the number line and marking point ,-5 and -7 on it.
Now the three intervals are (-∞ , -7], [-7,-5] and [-5,∞).
The solution set of inequality (x+5)(x+7)≥0 is (-∞ , -7] and [-5,∞).
The set of possible test points for
⇒ (-∞ , -7] → -8, -10
⇒ [-7,-5] → -6
⇒ [-5,∞) → -4, 0
Option (1) -8,-6,-4 and Option (2) -10,-6,0 satisfies the given condition.
