# Algebra

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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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#2

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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

**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

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(Original post by

Start with the lefthand side. Put it over a common denominator. What do you get?

**ghostwalker**)Start with the lefthand side. Put it over a common denominator. What do you get?

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**ghostwalker**)

Start with the lefthand side. Put it over a common denominator. What do you get?

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#5

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So it will be A(x+b)(x+c)+B(x+a)(x+c)+C(x+a)(x+b)= f(x)

**Shas72**)So it will be A(x+b)(x+c)+B(x+a)(x+c)+C(x+a)(x+b)= f(x)

(Original post by

Do I also take x=-a, x=-b and x=-c

**Shas72**)Do I also take x=-a, x=-b and x=-c

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Yes. So, expand and group terms in x^2, x,

**ghostwalker**)Yes. So, expand and group terms in x^2, x,

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

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#8

=(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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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?

**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?

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#10

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A+B+C

**Shas72**)A+B+C

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**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?

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#12

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So linear function is a straight line and is not quadratic. So x^2 doesnt exist. So A+B+C=0

**Shas72**)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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**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?

Can you also pls explain the reasoning whether it's possible to find the binomial expansion for sqrt(x-1)

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#14

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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)

**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)

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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(Original post by

Suppose

and further suppose you could write it as a power series.

Then

Fine if you're working with complex numbers,

**ghostwalker**)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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