The value of A, B & C

2007-11-07 6:20 pm
A (x-1)^2= B(x-1)(x-2)=9x^2-23x+C

Can any one tell me how to do this with 3 unknown ?

回答 (1)

2007-11-07 10:38 pm
✔ 最佳答案
A (x-1)^2= 9x^2-23x+C
A(x^2-2x+1)=9x^2-23x+C
Ax^2-2Ax+A=9x^2-23x+C
=> A=9, A=C=9
But, 2A=18 not equal to 23
So, for A (x-1)^2= 9x^2-23x+C, there is no A and C that satisfy this equation.

B(x-1)(x-2)=9x^2-23x+C
B(x^2-3x+2)=9x^2-23x+C
=> B=9, 2B=C=> C=18
But 3B=27 not equal to 23
So, for B(x-1)(x-2)=9x^2-23x+C, there is no B and C that satisfy this equation.

A (x-1)^2= B(x-1)(x-2)
Ax^2-2Ax+A=Bx^2-3Bx+2B
=> A=B
But 2A not equal to 3B
So, for A (x-1)^2= B(x-1)(x-2), there is no A and B that satisfy this equation.
參考: me


收錄日期: 2021-04-13 16:21:06
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20071107000051KK00653

檢視 Wayback Machine 備份