二次方程的根之和與積...

α+β= 2
α*β= 3/2

sum of new roots = (α^2- 1/β) + (β^2- 1/α)
= [α(α^2 - 1) + β(β^2- 1)] / (αβ)
= (α^3 + β^3 - α- β) / (αβ)
= [(α+β)(α^2 - αβ + β^2) - (α+β)] / (αβ)
= {(α+β)[(α+β)^2 - 3αβ] - (α+β)} / (αβ)
= {(2)[(2)^2 - 3*(3/2)] - (2)} / (3/2)
= -2

product of new roots = (α^2- 1/β)(β^2- 1/α)
= (α^2β^2 - α^2- β^2 + 1) / (αβ)
= [α^2β^2 - (α^2 + β^2) + 1] / (αβ)
= {α^2β^2 - [(α+β)^2 - 2αβ] + 1} / (αβ)
= 3/2

the required equation:
x^2 - (-2)x + (3/2) = 0
x^2 + 2x + 3/2 = 0
2x^2 + 4x + 3 = 0

已知α, β是二次方程2x^2-4x+3=0的根,試作一個關於x的二次方程,
其根為α^2- 1/β和β^2- 1/α
α+β= 2, αβ=3/2
(x-[α^2- 1/β])(x-[β^2- 1/α]) = 0
αβ(x-βα^2+1)(x-αβ^2+1) = 0
2(x-3α/2 + 1)(x-3β/2 + 1) = 0
(x-3α/2 + 1)(x-3β/2 + 1) = 0
x^2 + x(-3α/2 + 1-3β/2 + 1) + (3α/2 - 1)(3β/2 - 1) = 0
x^2 +x(-3[α+β]/2 + 2) + (9αβ/4 -3[α+β]/2 + 1) = 0
x^2 + x(-1) + (27/8 - 1) = 0
8x^2 - 8x + 19 = 0