代數問題(8)，數學達人請進。

(a) (2y-9)^2 = 5
2y-9 = ±√5
2y = √5 + 9 or √5 - 9
y = (√5 + 9)/2 or (√5 - 9)/2

(b) (3y+10)^2 + 81 = 0
(3y+10)^2 = -81
3y+10 = ±√-81
There is no real solution in this eqaution.

However, if you can express in terms of a+bi
3y+10 = ±9i
3y+10 = 9i or -9i
3y = 9i-10 or -9i-10
y = (9i-10)/3 or (-9i-10)/3

2010-09-16 16:26:03 補充：
Sorry for a mistake in (a)
2y = √5 + 9 or -√5 + 9
y = (√5 + 9)/2 or (9 - √5)/2

For checking,
(a) Put y = (√5 + 9)/2
= [2(√5 + 9)/2 -9]^2
= (√5)^2
= 5
Put y = (9 - √5)/2
= [2(9 - √5)/2 -9]^2
= (-√5)^2
= 5

2010-09-16 16:33:50 補充：
(b) Put y = (9i-10)/3
(3y+10)^2
= (9i-10+10)^2
= (9i)^2
= 81i^2
= 81√-1
= √-81
Put y = (-9i-10)/3
= (3y+10)^2
= (-9i)^2
= 81i^2
= √-81