F.4 Maths

α + β = 7 and αβ = -3
Let the required equation be x^2 + px + q = 0
So,
p = -(α/β + β/α)
= -(α^2 + β^2)/(αβ)
= -[(α + β)^2 - 2αβ]/(αβ)
= -(7^2 + 6)/(-3)
= 55/3
q = (α/β)(β/α) = 1
Therefore, the required equation is :
x^2 + 55x/3 + 1 = 0
==> 3x^2 + 55x + 3 = 0

首先，從原式 x² - 7x - 3 = 0 得到 α+β = 7 ; αβ= -3

假設以 α/β 和 β/α 為根的方程是 X² + hX + k = 0

α/β + β/α = (α² + β²)/αβ = -h
(α/β)(β/α) = 1 ; k = 1

(α² + β²)/αβ = -h

(α² + β²)/(-3) = -h

(α² + β²) = 3h

(α + β)² = α² + 2αβ + β² = 7² ; then (α² + β²) = .......