Past papers › JAMB 1992 · UME › Question 15 Question JAMB General Maths 1992 Objective Quadratics & their graphs Expressions, formulae & change of subject Quadratics & their graphs, Expressions, formulae & change of subject
Resolve 3 x 2 + x − 2 \dfrac{3}{x^2 + x - 2} x 2 + x − 2 3
A 1 x − 1 − 1 x + 2 \frac{1}{x - 1} - \frac{1}{x + 2} x − 1 1 − x + 2 1 B 1 x + 2 − 1 x − 1 \frac{1}{x + 2} - \frac{1}{x - 1} x + 2 1 − x − 1 1 C 1 x + 1 − 1 x − 2 \frac{1}{x + 1} - \frac{1}{x - 2} x + 1 1 − x − 2 1 D 1 x − 2 + 1 x + 1 \frac{1}{x - 2} + \frac{1}{x + 1} x − 2 1 + x + 1 1
Worked solution (try it first) Factorise the bottom:
x 2 + x − 2 = ( x − 1 ) ( x + 2 ) x^2 + x - 2 = (x - 1)(x + 2) x 2 + x − 2 = ( x − 1 ) ( x + 2 ) .
Write
3 ( x − 1 ) ( x + 2 ) = A x − 1 + B x + 2 \dfrac{3}{(x - 1)(x + 2)} = \dfrac{A}{x - 1} + \dfrac{B}{x + 2} ( x − 1 ) ( x + 2 ) 3 = x − 1 A + x + 2 B .
Multiply through by
( x − 1 ) ( x + 2 ) (x - 1)(x + 2) ( x − 1 ) ( x + 2 ) :
3 = A ( x + 2 ) + B ( x − 1 ) 3 = A(x + 2) + B(x - 1) 3 = A ( x + 2 ) + B ( x − 1 ) .
Put
x = 1 x = 1 x = 1 :
3 = 3 A 3 = 3A 3 = 3 A , so
A = 1 A = 1 A = 1 .
Put
x = − 2 x = -2 x = − 2 :
3 = − 3 B 3 = -3B 3 = − 3 B , so
B = − 1 B = -1 B = − 1 .
So the partial fractions are
1 x − 1 − 1 x + 2 \dfrac{1}{x - 1} - \dfrac{1}{x + 2} x − 1 1 − x + 2 1 , option A.
Watch out
Watch the order of the subtraction. Option B, 1 x + 2 − 1 x − 1 \frac{1}{x + 2} - \frac{1}{x - 1} x + 2 1 − x − 1 1 − 3 ( x − 1 ) ( x + 2 ) \frac{-3}{(x - 1)(x + 2)} ( x − 1 ) ( x + 2 ) − 3 Report a problem with this question