1) [(1 / ɑ) + (1 / √β)] and [(1 / ɑ) – (1 / √β)]
2) [(1 / √ɑ) + (1 / β)] and [(1 / √ɑ) – (1 / β)]
3) [(1 / √ɑ) + (1 / √β)] and [(1 / √ɑ) – (1 / √β)]
4) [(√ɑ) + (√β)] and [(√ɑ) – (√β)]
Solution: (1) [(1 / ɑ) + (1 / √β)] and [(1 / ɑ) – (1 / √β)]
(α + √β ) + (α – √β) = – p
The equation x2 + px + q = 0
α + √β + α – √β = – p; α2 – β = q
2α = – p; α2 – β = q
p2 / 4 – β = q
β = [p2 – 4q] / 4
[p2 – 4q] [p2x2 + 4px] – 16q = 04β (4α2x2 – 8αx) – 16 (α2 – β) = 0
β (4α2x2 – 8αx) – 4 (α2 – β) = 0
[(α √β x) – √β]2 = α2α √β x – √β = ± α
x = [(1 / ɑ) + (1 / √β)] and [(1 / ɑ) – (1 / √β)]