(a+1)^2 X^2 -3ax +(2a-1)=0
one of the roots of it is 1
a, what are the restrictions on the value of a?
B, find the value of a
2,kx^2 +(k-1)x+(k-1) =0 has a double real root, where k is a constant
A find the value of k
B, find the root of the equation when i) k>0 ii) k<0
maths
2007-07-23 7:27 pm
回答 (2)
2007-07-23 8:25 pm
✔ 最佳答案
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參考: My Maths knowledge
2007-07-23 8:38 pm
a.
a is not equal to -1
b.
The root is 1
(a+1)^2-3a+(2a-1)=0
a^2 +2a +1 -3a+2a-1=0
a^2+a=0
a(a+1)=0
a=0 or a =-1 (rejected by a)
2.the equation has double root.
Discriminant =0
(k-1)^2-4(k)(k-1)=0
k^2-2k+1-4k^2+4k=0
-3k^2+2k+1=0
3k^2-2k-1=0
(3k+1)(k-1)=0
k=-1/3 or k=1
i)
k=1
x^2=0
x=0
ii)
k=-1/3
-1/3x^2-4/3x-4/3=0
x^2+4x+4=0
(x+2)^2=0
x=-2
a is not equal to -1
b.
The root is 1
(a+1)^2-3a+(2a-1)=0
a^2 +2a +1 -3a+2a-1=0
a^2+a=0
a(a+1)=0
a=0 or a =-1 (rejected by a)
2.the equation has double root.
Discriminant =0
(k-1)^2-4(k)(k-1)=0
k^2-2k+1-4k^2+4k=0
-3k^2+2k+1=0
3k^2-2k-1=0
(3k+1)(k-1)=0
k=-1/3 or k=1
i)
k=1
x^2=0
x=0
ii)
k=-1/3
-1/3x^2-4/3x-4/3=0
x^2+4x+4=0
(x+2)^2=0
x=-2
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