Question 16

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Jambmaths question: 

The solution of the quadratic inequality is $({{x}^{2}}+x-12)\ge 0$is

Option A: 

$x\le 3\text{ or }x\le -4$

Option B: 

$x\ge 3\text{ or }x\le -4$

Option C: 

$x\ge -3\text{ or }x\le 4$

Option D: 

$x\ge 3\text{ or }x\ge -4$

Jamb Maths Solution: 

$({{x}^{2}}+x-12)\ge 0$

$(x+4)(x-3)\ge 0$

Find the range where this inequality is true.

 

$x\le -4$

$-4\le x\le 3$

$x\ge 3$

(x + 4 )

+

+

(x – 3 )

+

(x + 4 ) (x – 3 )

+

+

The range of range for which the inequality holds true is $x\le -4$ or $x\ge 3$

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