Mathematical Problem(Unanswered Questions)

1.
x² – 5x – 2000 = 0
x = [– (– 5) ± √((– 5)² – 4(1)(– 2000))]/(2(1))
= (5 ± √8025)/2
= (5 ± 5√321)/2
1/x = 2/(5 ± 5√321)
= [2(5 (– or +) 5√321)]/[(5 ± 5√321)(5 (– or +) 5√321)]
= (10 (– or +) 10√321)/(5² – (5√321)²)
= (10 (– or +) 10√321)/(– 8000)
= (– 1 ± √321)/800

∴(x – 2)² – (x – 1)² + 1/x – 2
= (x² – 4x + 4) – (x² – 2x + 1) + 1/x – 2
= 1 – 2x + 1/x
= 1 – 2((5 ± 5√321)/2) + (– 1 ± √321)/800
= (800 – 4000 (– or +) 4000√321 – 1 ± √321)/800
= (– 3201 (– or +) 3999√321)/800

2.
a^4 + 3a² – 1 = 0
a² = [– 3 ± √(3² – 4(1)(– 1))]/(2(1))
= (– 3 ± √13)/2

b² – 3b – 1 = 0
b = [– (– 3) ± √((– 3)² – 4(1)(– 1))]/(2(1))
= (3 ± √13)/2

∴a²b
= [(– 3 + √13)/2][(3 + √13)/2] or [(– 3 + √13)/2][(3 – √13)/2] or [(– 3 – √13)/2][(3 + √13)/2] or [(– 3 – √13)/2][(3 – √13)/2]
= (13 – 9)/4 or (– 9 + 6√13 – 13)/4 or (– 9 – 6√13 – 13)/4 or (13 – 9)/4
= 1 or (– 11 + 3√13)/2 or (– 11 – 3√13)/2 or 1

but a²b ≠ 1,
∴when a² = (– 3 + √13)/2 , b = (3 – √13)/2 , a²b = (– 11 + 3√13)/2
　when a² = (– 3 – √13)/2 , b = (3 + √13)/2 , a²b = (– 11 – 3√13)/2

b² – 3b – 1 = 0
b – 3 – 1/b = 0
b – 1/b = 3
(b – 1/b)² = 3²
b² – 2 + 1/b² = 9
b² + 1/b² = 11

∴a^4 b² + 1/b²
= (1 – 3a²)b² + 1/b²
= – 3a²b² + b² + 1/b²
= – 3a²(3b + 1) + b² + 1/b²
= – 9a²b – 3a² + 11
= – 9[(– 11 (– or +) 3√13)/2] – 3[(– 3 (– or +) √13)/2] + 11
= (99 ± 27√13 + 9 ± 3√13 + 22)/2
= 65 ± 15√13

3.
m³ + 2m² + 1997
= m(1 – m) + 2m² + 1997
= m – m² + 2m² + 1997
= m + m² + 1997
= 1 + 1997
= 1998