# Algebra

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#1
given that A/(x+a) +B/(x+b)+C/(x+c)=f(x)/[(x+a)(x+b)(x+c)], prove that if f(x) is linear then A+B+C=0
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10 months ago
#2
(Original post by Shas72)
given that A/(x+a) +B/(x+b)+C/(x+c)=f(x)/[(x+a)(x+b)(x+c)], prove that if f(x) is linear then A+B+C=0
Start with the lefthand side. Put it over a common denominator. What do you get?
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#3
(Original post by ghostwalker)
Start with the lefthand side. Put it over a common denominator. What do you get?
So it will be A(x+b)(x+c)+B(x+a)(x+c)+C(x+a)(x+b)= f(x)
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#4
(Original post by ghostwalker)
Start with the lefthand side. Put it over a common denominator. What do you get?
Do I also take x=-a, x=-b and x=-c
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10 months ago
#5
(Original post by Shas72)
So it will be A(x+b)(x+c)+B(x+a)(x+c)+C(x+a)(x+b)= f(x)
Yes. So, expand and group terms in x^2, x, and constants.

(Original post by Shas72)
Do I also take x=-a, x=-b and x=-c
No need.
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#6
(Original post by ghostwalker)
Yes. So, expand and group terms in x^2, x,
Ok let me try
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#7
(Original post by ghostwalker)
Yes. So, expand and group terms in x^2, x, and constants.

No need.
So I get
Ax^2+Bx^2+Cx^2+[A(b+c)+B(a+c)+C(a+b)]x + Abc +Bac +Cab
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10 months ago
#8
(Original post by Shas72)
So I get
Ax^2+Bx^2+Cx^2+[A(b+c)+B(a+c)+C(a+b)]x + Abc +Bac +Cab
Not checked the whole thing, so I assume it's correct.

=(A+B+C)x^2+[A(b+c)+B(a+c)+C(a+b)]x + (Abc +Bac +Cab)

OK. So, you're told it's a linear function. What's the coefficient of x^2 going to be then?
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#9
(Original post by ghostwalker)
Not checked the whole thing, so I assume it's correct.

=(A+B+C)x^2+[A(b+c)+B(a+c)+C(a+b)]x + (Abc +Bac +Cab)

OK. So, you're told it's a linear function. What's the coefficient of x^2 going to be then?
A+B+C
0
10 months ago
#10
(Original post by Shas72)
A+B+C
Yes, well, that's what it is in the function. But if the function is linear (i.e. not quadratic), what must that equal.
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#11
(Original post by ghostwalker)
Not checked the whole thing, so I assume it's correct.

=(A+B+C)x^2+[A(b+c)+B(a+c)+C(a+b)]x + (Abc +Bac +Cab)

OK. So, you're told it's a linear function. What's the coefficient of x^2 going to be then?
So linear function is a straight line and is not quadratic. So x^2 doesnt exist. So A+B+C=0
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10 months ago
#12
(Original post by Shas72)
So linear function is a straight line and is not quadratic. So x^2 doesnt exist. So A+B+C=0
Yep. Though we would say, the coefficient of x^2 must be zero, and therefore A+B+C=0,
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#13
(Original post by ghostwalker)
Not checked the whole thing, so I assume it's correct.

=(A+B+C)x^2+[A(b+c)+B(a+c)+C(a+b)]x + (Abc +Bac +Cab)

OK. So, you're told it's a linear function. What's the coefficient of x^2 going to be then?
Thank you so so much!
Can you also pls explain the reasoning whether it's possible to find the binomial expansion for sqrt(x-1)
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10 months ago
#14
(Original post by Shas72)
Thank you so so much!
Can you also pls explain the reasoning whether it's possible to find the binomial expansion for sqrt(x-1)
Suppose and further suppose you could write it as a power series. Then  Fine if you're working with complex numbers, but otherwise not.
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#15
Thank you very very much!!
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