解三次方程問題

2008-03-27 6:46 am

解三次方程 x^3=6x+40
(要求準確答案, 並須盡量簡化)
要求步驟, 唔準撞, 唔準用計數機, 唔接受近似值.
記住: 要盡量簡化

Solve the equation x^3=6x+40
(Give the exact answer in the simpliest form)
Steps are required.
You are asked not to use the guess-and-check method.
You should not use calculators.
Approximated value is not accepted.
Remember: Give the simpliest form

更新1:

Timothy, How do you know you should express -6x as -16x+10x but not -15x+9x or -100x+94x? ile0099, How do you factorize x^3-6x-40 as (x-4)(x^2+4x+10)?

更新2:

唔該解埋x^3=11x+14 (要求準確答案, 並須盡量簡化) 要求步驟, 唔準撞, 唔準用計數機, 唔接受近似值. 記住: 要盡量簡化

更新3:

Solve the equation x^3=11x+14 as well please (Give the exact answer in the simpliest form) Steps are required. You are asked not to use the guess-and-check method. You should not use calculators. Approximated value is not accepted. Remember: Give the simpliest form

更新4:

但係doraemonpaul, 你好似冇簡化.

更新5:

第2題答案應該係: -2, 1+2√2或1-2√2, 同你的答案好似有好大分別...

回答 (5)

doraemonpaul

2008-04-01 10:24 am

✔ 最佳答案

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慶哥

2008-03-27 8:44 am

其實三次函數用因式定理做係最快的,除非所有解都唔係整數
x^3=11x+14
撞x=正負1/正負2,都係撞四次
x^3-11x-14=0
(x-2)(x^2+2x-7)=0
....

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海獺雪

2008-03-27 7:21 am

x^3=6x+40
x^3-6x-40=0
(x-4)(x^2+4x+10)=0
∴x-4=0 or x^2+4x+10=0
x=4 or (rejected since △<0)

∴x=4

2008-03-26 23:21:56 補充：
△<0

2008-03-30 01:48:14 補充：
如果你問點樣可以知道因式分解
其實這個涉及一個繁複的步驟
有個方法可以幫到你

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設S= ax^3+bx^2+cx+d (其中a, b, c 及d 為常數, a 等於0)

即可改寫S= (a)[x^3 +(b/a)x^2 +(c/a)(x)+(d/a)].

失一般性，設S= x^3+bx^2+cx+d.

第一步，設y=x+(b/3)，即x=[y-(b/3)], ------(1)

可計算到S=y^3+{c-[(b)^2/3]}y+{d+[2(b)^3]/(27)- [(bc)]/3}

2008-03-30 01:48:44 補充：
第二步，設p={c-[(b)^2/3]} ----------(2)

及q={d+[2(b)^3]/(27)- [(bc)]/3} ----------(3)

即 S= y^3+py+q

再設 z 為 y=z- (p)/(3z) ------------- (4)

可計算到 S = (1/z^3){(z^3)^2+q(z^3)-[(p^3)/27]}。

第三步，設Z=z^3 -----------(5)

即S= (1/Z){Z^2+qZ-[(p^3)/27]}

2008-03-30 01:48:57 補充：
第四步，好了，若S=0, 即(1/Z){Z^2+qZ-[(p^3)/27]}=0，

即{Z^2+qZ-[(p^3)/27]}=0--------(6)

可以先從(2)及(3)算出p及q的值，從(6)算出Z的解(若有)，從(5)可知z的解，從(4)得y的解及最後從(1)得x的解。
------------

另外, 如果肯定係整數解
可以用trial and error既方法
以fator Thm.重覆代一些整數n入去f(x)度
在f(x)係一多項式時，如果f(n) = 0，即(x-n)係f(x)其中一個factor
之後用長除法找來因式分解條多項式

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2008-03-30 01:49:32 補充：
對不起, 因為要補充的實在太多
希望以上內容可以幫到你^^

參考: 自己, 書+自己+網上資源, 書+自己+網上資源, 書+自己+網上資源

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Timothy

2008-03-27 7:12 am

x^3=6x+40
x^3-6x-40=0
x^3-16x+10x-40=0
x(x^2-16)+10(x-4)=0
x(x-4)(x+4)+10(x-4)=0
[x(x+4)+10](x-4)=0
(x^2+4x+10)(x-4)=0
therefore
x^2+4x+10=0 or x-4=0
since b^2-4ac = 4^2-4(1)(10)=16-40=-24 , which is smaller than 0
therefore no real root for x^2+4x+10=0
i.e., x=4

參考: 試下咋...好耐冇做過呢d數, 希望幫到你

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A Maths student

2008-03-27 7:03 am

哈!題目是三次方程，又唔俾碰，又唔俾計數機，我都唔知點計=.=
最多只係試用remainder theorem,whenx=4,f(4)=0,so f(x) is divisible by(x-4)
可能有一個方法真係可以滿足題目要求：就係用三次公式

我以前都問過類似問題，不過佢地都只係俾答案

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