F.4 QUADRATIC EQUATION AND 其他數

2010-10-13 2:21 am

1, If a and b are the roots of the equation 2x^2+6x-3=0 and a>b, find the value of each of the following.

a^2-b^2

2. a^10+b^10 <---化簡

回答 (1)

知識就是力量

2010-10-13 2:41 am

✔ 最佳答案

1. a+b = -6/2 = -3
ab = -3/2

a^2 - b^2
= (a+b)(a-b)
= -3√(a-b)^2
= -3√(a^2 - 2ab + b^2)
= -3√[(a+b)^2 - 4ab]
= -3√[(-3)^2 - 4(-3/2)]
= -3√(9+6)
= -3√15

2. a^10 + b^10 (is already the simpliest form but you can factorize it)
= (a+b)(a^9 - a^8 b + a^7 b^2 - a^6 b^3 + a^5 b^4 - a^4 b^5 + a^3 b^6 - a^2 b^7 + ab^8 + b^9)

2010-10-12 21:45:41 補充：
(a+b)(a-b)
你知道 a+b = 3
但a-b如何處理
方法就是給它一個2次方
變成(a-b)^2 = a^2 - 2ab + b^2
但2次方不能白白的加所以要同時加一個"√"
√(a-b)^2 = a-b

參考: Knowledge is power.

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