一元二次方程

(k + 1)x² + (2k + 3)x + 2 =0
設一根為 β，則另一根為 -2β。

兩根之和：
β+ (-2β) = -(2k + 3)/(k + 1)
-β= -(2k + 3)/(k + 1)
β²= (2k + 3)²/(k + 1)² ...... [1]


兩根之積：
β•(-2β) = 2/(k + 1)
β² = -1/(k + 1) ...... [2]

[1] = [2] :
(2k + 3)²/(k + 1)² = -1/(k + 1)
(2k + 3)² = -(k + 1)
4k² + 12k + 9 = -k - 1
4k² + 13k + 10 = 0
(4k + 5)(k + 2) = 0
k = -5/4 或 k = -2