Question

X²-5X+4=0

Answer 1

更新1:

(k-3)x² - 5x + 4 = 0 change the question... the correct question should be this...

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 补充： 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 补充： (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 补充： Correction: (k-3)x² - 5x + 4 = 0 x = [5+/-√(25-16(k-3))]/2(k-3) x = [5+/-√(73-16k)]/2(k-3)
参考: Hope the solution can help you^^”
(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