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