Jambmaths question:
If $y={{x}^{2}}-x-12,$find the range of x for which y ≥ 0
Option A:
$x<-3\text{ or }x>4$
Option B:
$x\le -3\text{ or }x\ge 4$
Option C:
$-3<x\le 4$
Option D:
$-3\le x\le 4$
Jamb Maths Solution:
${{x}^{2}}-x-12\ge 0$
$(x-4)(x+3)\ge 0$
|
|
$x\le -3$ |
$-3\le x\le 4$ |
$x\ge 4$ |
|
x – 4 |
– |
– |
+ |
|
x + 3 |
– |
+ |
+ |
|
(x – 4)( x + 3) |
+ |
– |
+ |
From the range table, the value(s) for which x y ≥ 0 are $x\le -3$ and $x\ge 4$
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