The inequality x2 12x 35 ≥ 0 has two critical points and three possible intervals for solutions.
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 on drawing the number line and marking point ,-5 and -7 on it, we get the following three intervals, (-∞ , -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 satisfy the given condition.
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