X²-5X+4=0

x^2 - 5x + 4 = 0
x^2 + (-4-1)x + (-4*-1) = 0
(x-4)(x-1) = 0
x = 4 or 1


2011-10-08 19:56:46 補充：
Alternative method,
By x = [-b+/-√(b^2 - 4ac)]/2a for ax^2 + bx + c = 0
In the equation,
a = 1, b = -5, c = 4
x = [5+/-√(25-16)]/2 = (5+/-3)/2 = 8/2 or 2/2 = 1 or 4

2011-10-08 20:10:08 補充：
(k-3)x² - 5x + 4 = 0
x = [5+/-√(24-16(k-3))]/2(k-3)
x = [5+/-√(72-16k)]/2(k-3)
x = [5+/-2√(18-4k)]/2(k-3)

2011-10-08 23:30:19 補充：
Correction:
(k-3)x² - 5x + 4 = 0
x = [5+/-√(25-16(k-3))]/2(k-3)
x = [5+/-√(73-16k)]/2(k-3)

(k-3)x^2 - 5x + 4 = 0
x=(5+√((k-3)^2-4(k-3)(4))/2(k+3) or (5-√((k-3)^2-4(k-3)(4))/2(k+3)
=(5+√(k^2-6k+9-16k+12))/(2k+6) or (5+√(k^2-6k+9-16k+12))/(2k+6)
=(5+√(k-21)(k-1))/2k+6 or (5+√(k-21)(k-1))/2k+6