高中數學二次方程問題求救?

For x²+2(a-1)x-(a²+4a-6ab+9b²)=0 :
discriminant, ∆=[2(a-1)]²+4(a²+4a-6ab+9b²)

If the eqt. has 1 double real root,
4(a-1)²+4(a²+4a-6ab+9b²)=0
(a-1)²+(a²+4a-6ab+9b²)=0
a²-2a+1+a²+4a-6ab+9b²=0
(a²+2a+1)+(a²-6ab+9b²)=0 ←---------- (#)
(a+1)²+(a-3b)²=0
(a+1)²=0 and (a-3b)²=0
a= -1 and a-3b=0
a= -1 and (-1)-3b=0
∴ a= -1 and b= - ⅓

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(#) :
這一行的構思 , 是由 " (a) part " 給出的提示而來！