factorise this: x^2-4x+5=0?
回答 (5)
2016-06-20 7:59 am
Method 1 :
x² - 4x + 5 = 0
x² - 4x = -5
x² - 4x + 4 = -5 + 4
(x - 2)² = -1
x - 2 = √-1 or x - 2 = -√-1
x - 2 = i or x - 2 = -i
x = 2 + i or x = 2 - i
(where i = √-1)
====
Method 2 :
If x² + bx + c = 0, then x = [-b ± √(b² - 4c)]/2
Now, x² - 4x + 5 = 0
x = {-(-4) ± √[(-4)² - 4(5)]}/2
x = (4 ± √-4)/2
x = (4 ± 2i)/2
x = 2 + i or x = 2 - i
(where i = √-1)
x² - 4x + 5 = 0
x² - 4x = -5
x² - 4x + 4 = -5 + 4
(x - 2)² = -1
x - 2 = √-1 or x - 2 = -√-1
x - 2 = i or x - 2 = -i
x = 2 + i or x = 2 - i
(where i = √-1)
====
Method 2 :
If x² + bx + c = 0, then x = [-b ± √(b² - 4c)]/2
Now, x² - 4x + 5 = 0
x = {-(-4) ± √[(-4)² - 4(5)]}/2
x = (4 ± √-4)/2
x = (4 ± 2i)/2
x = 2 + i or x = 2 - i
(where i = √-1)
2016-06-20 8:59 am
x^2 - 4x + 5 = 0
(x - 2)^2 + 1 = 0
Complex solutions:
x = 2 - i
x = 2 + i
(x - 2)^2 + 1 = 0
Complex solutions:
x = 2 - i
x = 2 + i
2016-06-20 8:59 am
Still does not factorise !!!
2016-06-20 8:01 am
x² - 4x + 5 = 0
x² - 4x + 4 + 1 = 0
(x² - 4x + 4) + 1 = 0
(x - 2)² + 1 = 0
(x - 2)² - i² = 0 → you recognize: a² - b² = (a + b).(a - b)
[(x - 2) + i].[(x - 2) - i] = 0
(x - 2 + i).(x - 2 - i) = 0
x² - 4x + 4 + 1 = 0
(x² - 4x + 4) + 1 = 0
(x - 2)² + 1 = 0
(x - 2)² - i² = 0 → you recognize: a² - b² = (a + b).(a - b)
[(x - 2) + i].[(x - 2) - i] = 0
(x - 2 + i).(x - 2 - i) = 0
2016-06-20 7:57 am
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