Past papers › WAEC · 2024 · May/June · General Maths · Paper 1 › Question 50 Question WAEC General Maths 2024 Objective Expressions, formulae & change of subject Quadratics & their graphs Expressions, formulae & change of subject, Quadratics & their graphs
WAEC 2024 · Paper 1 · Q50 For what values of y y y y + 2 8 y 2 − 10 y + 3 \dfrac{y + 2}{8y^2 - 10y + 3} 8 y 2 − 10 y + 3 y + 2
A − 3 4 , 1 2 -\frac34, \frac12 − 4 3 , 2 1 B − 3 4 , − 1 2 -\frac34, -\frac12 − 4 3 , − 2 1 C 3 4 , 1 2 \frac34, \frac12 4 3 , 2 1 D 3 4 , − 1 2 \frac34, -\frac12 4 3 , − 2 1
Worked solution (try it first) The fraction is not defined when the bottom is zero:
8 y 2 − 10 y + 3 = 0 8y^2 - 10y + 3 = 0 8 y 2 − 10 y + 3 = 0 .
Factorise (numbers
− 6 -6 − 6 and
− 4 -4 − 4 multiply to 24 and add to
− 10 -10 − 10 ):
( 4 y − 3 ) ( 2 y − 1 ) = 0 (4y - 3)(2y - 1) = 0 ( 4 y − 3 ) ( 2 y − 1 ) = 0 .
So
y = 3 4 y = \frac34 y = 4 3 or
y = 1 2 y = \frac12 y = 2 1 , option C.
Watch out
Both brackets have a minus, so both roots are positive. A sign slip in either bracket gives a negative root, as in options A, B and D. Report a problem with this question