Question 33

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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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