用配方法求方程 aX2+abX－2=0 (a＞0)的解.

aX2+abX－2=0,
a(x^2+bx)-2=0,
a[x^2+bx+(b/2)^2-(b/2)^2]-2=0,
a[x+(b/2)]^2-(ab^2/4)-2=0,
a[x+(b/2)]^2=(ab^2+8)/4,
[x+(b/2)]^2=(ab^2+8)/(4a),
x+(b/2)=±√[(ab^2+8)/(4a)]
x=-(b/2)±√[(ab^2+8)/(4a)]
=(-b/2)±√(a^2b^2+8a)/(2a)
=[-ab±√(a^2b^2+8a)]/(2a).

ax^2+abx -2 =0
a[x^2+bx + (b/2)^2] = 2 + ab^2/4
a( x+ b/2) ^2 = (8+ab^2)/4
( x+ b/2) ^2 = (8+ab^2)/(4a)
x+b/2 = ±√{(8+ab^2)/(4a)}
x =-b/2 ±√{(8+ab^2)/(4a)}

a(X+(ab-[(ab)^2+8a])/2ab)(X+(ab+[(ab)^2+8a])/2ab)=0
所以X1=(-ab+[(ab)^2+8a])/2ab
X2=(-ab-[(ab)^2+8a])/2ab
注意:[]内为根号下的