F.4 Quadratic equation~! HARD!

(1a) α^2 + β^2 = 21 ... (1)
α^2β^2 = 4 => αβ = 2 ...(2)
(2) => α^2 + 2αβ + β^2 - 2αβ = 21
(α + β)^2 - 4 = 21
(α + β)^2 = 25
α + β = 5
The equαtion is x^2 - 5x + 2 = 0
(b) (i) (α - β)^2 = α^2 - 2αβ + β^2
= α^2 + 2αβ + β^2 - 4αβ
= (α + β)^2 - 4αβ
= 25 - 8
= 17
α - β = -√17
(ii) Since α αnd β αre roots of x^2 - 5x + 2 = 0
α^2 - 5α + 2 = 0
α^2 = 5α - 2
α^2 - 5β = 5α - 2 - 5β = -2 + 5(α - β) = -2 - 5√17
(iii) α^3β - 2(2 - β)(β + 2)
= α^2(αβ) + 2(β - 2)(β + 2)
= 2α^2 + 2β^2 - 8
= 2(α^2 + β^2) - 8
= 42 - 8
= 34
(iv) (β^2 + 2)/2β + √(17α^2) + β^2
= (5β - 2 + 2)/2β + (√17)α + 5β - 2
= 5/2 - 2 + (√17)α + 5β
= 1/2 + (√17 + 5)/2 (α + β) + (√17 - 5)/2 (α - β)
= 1/2 + 5(√17 + 5)/2 - √17(√17 - 5)/2
= 1/2 + (5√17 + 25)/2 - (17 - 5√17)/2
= 1/2 + 4 + 5√17
= 9/2 + 5√17
(2) (a) f(x) = ax^2 + bx + c
f(x) = a(x^2 + bx/a) + c
f(x) = a(x + b/2a)^2 + c - b^2/4a
f(x) = a(x + b/2a)^2 - (b^2 - 4ac)/4a
The vertex is [-b/2a, -(b^2 - 4ac)/4a]
(b) y = f(x)/2 = (ax^2 + bx + c)/2
When x = 0, f(0)/2 = c/2 = 1.5
c = 3
(c)(i) vertex of f(x)/2 is [-b/2a, -(b^2 - 4ac)/8a]
So -(b^2 - 4ac) / 8a = 25/16
-(b^2 - 12a) / 8a = 25/16
-b^2 + 12a = 25a/2
-2b^2 + 24a = 25a
a = -2b^2
(ii) f(x) = -2b^2x^2 + bx + 3
Sub(2, -7) into equation,
-7 = -2b^2(4) + 2b + 3
-8b^2 + 2b + 10 = 0
4b^2 - b - 5 = 0
(4b - 5)(b + 1) = 0
b = 5/4 or b = -1
When b = -1,
a = -2 αnd d = -b/2a = -1/4
when b = 5/4,
a = (-2)(25/16) = -25/8
d = -b/2a = -(5/4)(-4/25) = 5(rejected)
So a = -2, b = -1 αnd d = -1/4
(d) The axis of symmetry for both (*) and (**) are the same as x = d
i.e. x = -1/4
(e) f(x) = -2x^2 - x + 3 = 0
2x^2 + x - 3 = 0
(2x + 3)(x - 1) = 0
x = 1 or x = -3/2