Maths Q

2007-07-27 7:25 pm
Find the values of A and B for the following identities in x:

1. (A-B)x+(A+B) identity to 7x+3


2. x square+Ax+10 identity to (x-2)(x-B)

回答 (4)

2007-07-27 7:37 pm
✔ 最佳答案
A-B = 7
A+B = 3
so, 2A=10
A=5
B= -2

(X-2)(X-B) = X^2 + (-2-B)X + 2B
So, A = -2-B
2B = 10
B=5
A= -7
2007-07-27 7:50 pm

圖片參考:http://hk.yimg.com/i/icon/16/1.gif

(A-B)=7...(1
(A+B)=3...(2
(1+(2=A-B+A+B=7+3
2A=10
圖片參考:http://hk.yimg.com/i/icon/16/58.gif

A=5,代A=5進A-B=7 因此B=-2
檢驗代進2式5+(-2)=3,成立
圖片參考:http://hk.yimg.com/i/icon/16/67.gif

2. x square+Ax+10 identity to (x-2)(x-B) ...1)
圖片參考:http://hk.yimg.com/i/icon/16/58.gif

先把(X-2)(X-B)拆散=Xsquare +(-2-B)X+2B
可以看到2B=10和+AX=(-2-B)X
因此B=5
A=(-2-B)
A=-7
代進A=-7和B=5進1)
x square-7x+10恆等於 (x-2)(x-5),成立
圖片參考:http://hk.yimg.com/i/icon/16/67.gif


2007-07-27 19:53:28 補充:
Q2. x square Ax=10 identity to (x-2)(x B) 先把(X-2)(X+B)拆散=Xsquare +(B-2)X-2Bx square Ax=10 ,可以寫成x square Ax-10=0因此-10=-2BB=5由於Ax=(B-2)x  A=(B-2)  A=5-2  A=3代進A=3和B=5進問題x square+3X-10=(X-2)(X+5),因此恆等式成立

2007-07-27 19:56:00 補充:
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2007-07-30 12:05:21 補充:
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2007-09-07 15:57:17 補充:
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2007-07-27 7:42 pm
1)
A - B = 7-------------(1)
A + B = 3--------------(2)
Add (1) to (2)
2A = 10
a = 5
Substitute A = 5 into (1)
5 - B = 7
B = -2

2)
x^2 + Ax + 10 identity to (x-2)(x-B)

(x-2)(x-B) = x^2 - (B + 2)x + 2B identity to x^2 + Ax + 10
By conparing the coefficient
A = (-B) - 2 --------------(1)
2B = 10---------------(2)
Solve (2)
B = 5
Substitute B = 5 into (1)
A = (-5) - 2 = -7
2007-07-27 7:38 pm
1.
A - B = 7 ------------(1)
A + B = 3 -----------(2)
(1)+(2)
2A =10
A=5
put A=5 into (2)
5 + B =3
B = -2


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