急 ! F4 Quadratic Equations 1

2011-09-17 5:30 am
請詳細教我計以下幾條 :

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回答 (1)

2011-09-17 6:42 am
✔ 最佳答案
13a.
(y-2)^2 = 6
y - 2 = √6 or -√6
y = 2+√6 or 2-√6
b.
The new roots = 3+√6 and 3-√6
New equation:
[y-(3+√6)][y-(3-√6)] = 0
y^2 - (3+√6+3-√6)y + (3+√6)(3-√6) = 0
y^2 + 6y + 3 = 0
14a.
4x^2 - 8x + 3 = 0
x = [-(-8)+/-√(64-48)]/8
x = 1 +/- 1/2
b.
New roots
= (1+1/2 +1 -1/2) and (1+1/2)(1-1/2)
= 2 and 3/4
The new equation:
(y-2)(y-3/4) = 0
y^2 - 11y/4 + 6/4 = 0
4y^2 - 11y + 6 = 0
15a.
(x-a)(x-b) = 0
x^2 - (a+b)x + ab = 0
b.
ab = 12 and a, b=/=0
(a,b) = (+/-1,+/-12),(+/-2,+/-6),(+/-3,+/-4),(+/-4,+/-3),(+/-6,+/-2),(+/-12,+/-1)
參考: Hope the solution can help you^^”


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