F4 Maths

2013-05-23 11:07 pm
(1) x(3x+1) =0

(2) 3x^2 -6x+2 =0

(3) x^2-4x =0

(4) x^2-8X+12 =0

(5) x^2-8X+16 =0



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回答 (2)

2013-05-24 12:37 am
✔ 最佳答案
(1)
x(3x + 1) = 0
x = 0 or 3x + 1 = 0
x = 0 or x = -1/3


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(2)
3x² - 6x + 2 = 0
3(x² - 2x) = -2
3(x² - 2x + 1) = -2 + 3
3(x - 1)² = 1
9(x - 1)² = 3
3(x - 1) = √3 or 3(x - 1) = -√3
x - 1 = (√3)/3 or x - 1 = -(√3)/3
x = (3 + √3)/3 or x = (3 - √3)/3
(OR: Use the formula x = [-b ± √(b² - 4ac)] / 2a)


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(3)
x² - 4x =0
x(x - 4) = 0
x = 0 or x - 4 = 0
x = 0 or x = 4


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(4)
x² -8x + 12 =0
(x - 6)(x - 2) = 0
x - 6 = 0 or x - 2 = 0
x = 6 or x = 2


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(5)
x² - 8x + 16 =0
x² - 2x(4) + 4² = 0
(x - 4)² = 0
x = 4 (double roots)
參考: andrew
2013-05-23 11:16 pm
x(3x+1)=0
thus x = 0 or x = -1/3

3x^2 - 6x +2 = 0
x = ±(6sqrt(36-4*3*2))/6
=±6sqrt(12)/6
=±2sqrt(3)

x^2-4x=0
x(x-4)=0
x=4 or 0

x^2-8x+12=0
(x-6)(x-2)=0
x = 6 or 2

x^2-8x+16
(x-4)^2 =0
x=4


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