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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