問幾題數 - 配方法 (急)

2008-10-03 1:45 am
我想問3題數啊 唔該=]

用配方法解下列各二次方程

1. 2x^2-10x-5 = 0

2. 3x^2+7x-3 = 0

3. 5(x^2 -1) = 4x


唔該晒!!thank you!!!!!

回答 (3)

2008-10-03 2:11 am
✔ 最佳答案
1.
2x^2-10x-5 = 0
x^2-5x-5/2=0
x^2-5x+(5/2)^2-(5/2)^2-5/2=0
(x-5/2)^2-35/4=0
[x-5/2+√(35/4)][x-5/2-√(35/4)]=0
x=(5-√35)/2 or x=(5+√35)/2

2.
3x^2+7x-3 = 0
x^2+7/3x-1 = 0
x^2+7/3x+(7/6)^2-(7/6)^2-1 = 0
(x+7/6)^2-85/36=0
(x+7/6+√85/6)(x+7/6-√85/6)=0
x= (±√85-7)/6

3.
5(x^2 -1) = 4x
5x^2-4x-5=0
x^2-4/5x-1=0
x^2-4/5x+(4/10)^2-(4/10)^2-1=0
(x-2/5)^2-29/25=0
(x-2/5+ √29/5)(x-2/5- √29/5)=0
x=(2±√29)/5
2008-10-03 2:58 am
1. 2x^2 - 10x - 5 = 0

x^2 - 2(5/2)x + (5/2)^2 - (5/2)^2 - 5/2 = 0

(x - 5/2)^2 - (5/2)(5/2 + 1) = 0

(x - 5/2)^2 - (5/2)(7/2) = 0

(x - 5/2)^2 - 35/4 = 0

(x - 5/2)^2 - (√35/√4)^2 = 0

(x - 5/2 + √35/2)(x - 5/2 - √35/2) = 0

∴ x = (5 - √35)/2 or x = (5 + √35)/2


2. 3x^2 + 7x - 3 = 0

x^2 + (7/3)x - 1 = 0

x^2 + 2(7/3)(1/2)x + (7/6)^2 - (7/6)^2 - 1 = 0

(x + 7/6)^2 - ((49/36) + 1) = 0

(x + 7/6)^2 - 85/36 = 0

(x + 7/6)^2 - (√85/√36)^2 = 0

(x + 7/6)^2 - (√85/6)^2 = 0

(x + 7/6 + √85/6)(x + 7/6 - √85/6) = 0

∴ x = - (7 + √85)/6 or x = - (7 - √85)/6


3. 5(x^2 -1) = 4x

x^2 - 1 = (4/5)x

x^2 - (4/5)x - 1 = 0

x^2 - 2(1/2)(4/5)x + (4/10)^2 - (4/10)^2 - 1 = 0

(x - 4/10)^2 - 29/25 = 0

(x - 2/5)^2 - (√29/√25)^2 = 0

(x -2/5 + (√29/√25))(x - 2/5 - (√29/√25)) = 0

(x - (2 - √29)/5)(x - (2 + √29)/5) = 0

∴ x = (2 - √29)/5 or x = (2 + √29)/5
參考: My own calculation
2008-10-03 2:51 am
1.
2x^2-10x-5 = 0
x^2-5x-5/2=0
x^2-5x+(5/2)^2-(5/2)^2-5/2=0
(x-5/2)^2-35/4=0
[x-5/2+√(35/4)][x-5/2-√(35/4)]=0
x=(5-√35)/2 or x=(5+√35)/2

2.
3x^2+7x-3 = 0
x^2+7/3x-1 = 0
x^2+7/3x+(7/6)^2-(7/6)^2-1 = 0
(x+7/6)^2-85/36=0
(x+7/6+√85/6)(x+7/6-√85/6)=0
x= (±√85-7)/6

3.
5(x^2 -1) = 4x
5x^2-4x-5=0
x^2-4/5x-1=0
x^2-4/5x+(4/10)^2-(4/10)^2-1=0
(x-2/5)^2-29/25=0
(x-2/5+ √29/5)(x-2/5- √29/5)=0
x=(2±√29)/5 有步驟^_^


收錄日期: 2021-04-15 18:17:45
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20081002000051KK01192

檢視 Wayback Machine 備份