中四A-maths,二次方程的根之和與積,要列式(10分)高手教我吧！

1. α+ β = 3, αβ = 1
(a) sum of new roots = (4α-3) + (4β-3) = 4(α+ β) - 6 = 4(3) - 6 = 6
product of new roots = (4α-3)(4β-3) = 16αβ - 12(α+ β) + 9 = 16(1) - 12(3) + 9 = -11
the new equation : x^2 - (sum of new roots) + (product of new roots) = 0
x^2 - 6x - 11 = 0
(b) sum of new roots=2/α^2 + 2/β^2 = [2(α^2 + β^2]/[α^2β^2] = (2[(α+ β)^2 - 2αβ])/(αβ)^2
= (2[(3)^2 - 2(1)]) / (1)^2 = 14
product of new roots = 2/α^2 * 2/β^2 = 4 / (αβ)^2 = 4 / (1)^2 = 4
the new equation : x^2 - 14x + 4 = 0
(c) sum of new roots = (3α^2+2) + (3β^2+2) = 3[(α+ β)^2 - 2αβ] + 4 = 3[(3)^2 - 2(1)] + 4
= 25
product of new roots = (3α^2+2)(3β^2+2) = 9(αβ)^2 + 6[(α+ β)^2 - 2αβ] + 4
= 9(1)^2 + 6[(3)^2 - 2(1)] + 4 = 55

2. α+ β = -2.5, αβ = -2
(a) (α-α/β)(β-β/α) = αβ - β - α + 1 = αβ - (β + α) + 1 = -2 - (-2.5) + 1 = 1.5
(b) 2α^2+2β^2 = 2(α^2+β^2) = 2[(α+ β)^2 - 2αβ] = 2[(-2.5)^2 - 2(-2)] = 20.5
(c) (3α+β)(α+3β) = 3α^2 + 9αβ + αβ + 3β^2 = 3(α^2+β^2) + 10αβ
= 3[(α+ β)^2 - 2αβ] + 10αβ = 3[(-2.5)^2 - 2(-2)] + 10(-2) = 10.75
(d) α^3β+αβ^3 = αβ(α^2 + β^2) = αβ[(α+ β)^2 - 2αβ] = (-2)*[(-2.5)^2 - 2(-2)] = -20.5