Past papers › NECO · 2022 · SSCE · General Maths · Paper 1 › Question 19 Question NECO General Maths 2022 Objective Expressions, formulae & change of subject Quadratics & their graphs Expressions, formulae & change of subject, Quadratics & their graphs
NECO 2022 · Paper 1 · Q19 Find the values of x x x 3 x 2 + 6 x − 2 ( x + 2 ) ( x − 4 ) \dfrac{3x^2 + 6x - 2}{(x + 2)(x - 4)} ( x + 2 ) ( x − 4 ) 3 x 2 + 6 x − 2
A x = 2 x = 2 x = 2 4 4 4 B x = 2 x = 2 x = 2 − 4 -4 − 4 C x = − 2 x = -2 x = − 2 4 4 4 D x = 0 x = 0 x = 0 4 4 4 E x = 0 x = 0 x = 0 2 2 2
Worked solution (try it first) A fraction is undefined when its denominator is zero:
( x + 2 ) ( x − 4 ) = 0 (x + 2)(x - 4) = 0 ( x + 2 ) ( x − 4 ) = 0 .
So
x + 2 = 0 x + 2 = 0 x + 2 = 0 or
x − 4 = 0 x - 4 = 0 x − 4 = 0 , giving
x = − 2 x = -2 x = − 2 or
x = 4 x = 4 x = 4 , option C.
Watch out
Each factor gives the value that makes it zero, so x + 2 x + 2 x + 2 x = − 2 x = -2 x = − 2 x = 2 x = 2 x = 2 − 4 -4 − 4 Report a problem with this question