Past papers › JAMB 1992 · UME › Question 10 Question JAMB General Maths 1992 Objective Quadratics & their graphs Expressions, formulae & change of subject Quadratics & their graphs, Expressions, formulae & change of subject
Make t t t s = u t + 1 2 a t 2 s = ut + \frac12at^2 s = u t + 2 1 a t 2
A 1 a [ u ± u 2 − 2 a s ] \frac1a[u \pm \sqrt{u^2 - 2as}] a 1 [ u ± u 2 − 2 a s ] B 1 a [ − u ± u 2 − 2 a s ] \frac1a[-u \pm \sqrt{u^2 - 2as}] a 1 [ − u ± u 2 − 2 a s ] C 1 a [ u ± u 2 + 2 a s ] \frac1a[u \pm \sqrt{u^2 + 2as}] a 1 [ u ± u 2 + 2 a s ] D 1 a [ − u ± u 2 + 2 a s ] \frac1a[-u \pm \sqrt{u^2 + 2as}] a 1 [ − u ± u 2 + 2 a s ]
Worked solution (try it first) Multiply by 2 and bring everything to one side:
a t 2 + 2 u t − 2 s = 0 at^2 + 2ut - 2s = 0 a t 2 + 2 u t − 2 s = 0 , a quadratic in
t t t .
Use the formula with
a a a ,
b = 2 u b = 2u b = 2 u and
c = − 2 s c = -2s c = − 2 s :
t = − 2 u ± 4 u 2 + 8 a s 2 a t = \dfrac{-2u \pm \sqrt{4u^2 + 8as}}{2a} t = 2 a − 2 u ± 4 u 2 + 8 a s .
Take 4 out of the square root, which gives
2 u 2 + 2 a s 2\sqrt{u^2 + 2as} 2 u 2 + 2 a s , and divide top and bottom by 2.
So
t = 1 a [ − u ± u 2 + 2 a s ] t = \frac1a[-u \pm \sqrt{u^2 + 2as}] t = a 1 [ − u ± u 2 + 2 a s ] , option D.
Watch out
Here c = − 2 s c = -2s c = − 2 s − 4 a c = + 8 a s -4ac = +8as − 4 a c = + 8 a s u 2 + 2 a s u^2 + 2as u 2 + 2 a s − 8 a s -8as − 8 a s Report a problem with this question