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Quadratic Equations, SPM Practice (Paper 2)

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Question 2:
Given α and β are two roots of the quadratic equation (2x + 5)(x + 1) + p = 0 where αβ = 3 and p is a constant.
Find the value p, α and of β.

Solutions:
(2x + 5)(x + 1) + p = 0
2x2 + 2x + 5x + 5 + p = 0
2x2 + 7x + 5 + p = 0
*Compare with, x2– (sum of roots)x + product of roots = 0
x 2 + 7 2 x + 5 + p 2 = 0 ← divide both  sides with 2
Product of roots, αβ = 3
5 + p 2 = 3  
5 + p = 6
p = 1

Sum of roots = − 7 2  
   α + β = − 7 2   →  (1) and  α β = 3    →  (2) from (2),  β = 3 α    →  (3) Substitute (3) into (1), α + 3 α = − 7 2  

2α2+ 6 = –7α ← (multiply both sides with 2α)
2α2+ 7α + 6 = 0
(2α + 3)(α + 2) = 0
2α + 3 = 0   or α + 2 = 0
α=− 3 2    α = –2

Substitute  α = − 3 2  into (3), β = 3 − 3 2 = 3 ( − 2 3 ) = − 2

Substitute α = –2 into (3),
β = − 3 2 ∴   p = 1 ,  and when  α = − 3 2 , β = − 2  and  α = − 2 , β = − 3 2 .