複數計算請幫忙

a. 5x^ - 6x + 5=0 let it be 5x2 - 6x + 5=0

5x2 - 6x + 5=0

X = [-b±Ö(b^2 - 4ac)]/2a

X =[6±Ö(36-4(5)(5)]/2(5)

X = [6±Ö(-64)]/10

X = 6/10 ± (8/10)i

X = 3/5 + (4/5)i or 3/5 - (4/5)i


b: a=3-2j and b=-1+j

The required equation is [X-(3-2j)][X-(-1+j)] = 0

I am not quite sure about your question. I think this is what you want.
In electronics, i is used for current. j represent imaginary number i.

(X-a)(X-b)=0

X^2 - (a+b)X +ab = 0

X^2 -[(3-2j)+(-1+j)]X + (3-2j)(-1+j) = 0

X^2 -(2-3j)X + (-3+3j+2j+2) = 0

X^2 -(2-3j)X + 1+5j = 0




2013-09-28 12:00:52 補充：
Something goes wrong 5x^ - 6x + 5=0 let it be 5x^2 - 6x + 5=0

Ö is square root symbol

2013-09-28 12:06:59 補充：
c: if x = 3+4j is a root of x^+ax + b = 0 find the real numbers a and b

This is a property of quadratic equation; while one root is p + qi, the other root is p - qi.

The required equation is [X-(3+4j)][X-(3-4j)] = 0

2013-09-28 12:09:55 補充：
(3+4j)+(3-4j) = 6 = a

(3+4j)(3-4j) = 9 - 12j +12j +16 = 25 = b