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Find all values of A, B and C.

Could someone teach how to do this?

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- #1

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Find all values of A, B and C.

Could someone teach how to do this?

- #2

arildno

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Rewrite

[tex]x^{2}=((x+1)-1)^{2}[/tex]

and do the necessary operations.

[tex]x^{2}=((x+1)-1)^{2}[/tex]

and do the necessary operations.

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Huh? Sorry I don't understand what u mean.

- #4

arildno

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[tex]((x+1)-1)^{2}=(x+1)^{2}-2(x+1)+1[/tex]

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related to [tex]3x^2 + 4x + C \equiv A(x + 1)^2 + B(x + 1) + 7 ?[/tex]

Sorry but i don't get u.

- #6

arildno

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Make a similar rewriting of x:

[tex]x=((x+1)-1)[/tex]

Now, substitute these expressions for [tex]x,x^{2}[/tex] into your LEFT-HAND SIDE.

Reorganize the terms you get, and derive conditions so that your new expression equals your ORIGINAL RIGHT-HAND SIDE.

This will determine A,B,C.

- #7

Hurkyl

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Or you could plug in a few values for x to generate equations.

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- #9

arildno

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So (x+1)-1=x+1-1=x.

To be nice, I'll do this for once:

We have:

[tex]3x^{2}=3((x+1)-1)^{2}=3(x+1)^{2}-6(x+1)+3[/tex]

[tex]4x=4((x+1)-1)=4(x+1)-4[/tex]

[tex]C=C=C[/tex]

Now, add these equations together, downwards. The outermost terms then turn into:

[tex]3x^{2}+4x+C=3(x+1)^{2}-2(x+1)+(C-1)[/tex]

Do you understand this?

- #10

Hurkyl

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What does (x + 1) - 1 equal? Then what does ((x+1) - 1)^2 equal?

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I'm just curious, has this got something to do with modulus?

- #12

arildno

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Not that I know of..Sariaht said:I'm just curious, has this got something to do with modulus?

It is simply to substitute "equal for equal"

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<Sarcastic mode ON>

arildno you have to be a little more diplomatic, I think

<Sarcastic mode OFF>

arildno you have to be a little more diplomatic, I think

<Sarcastic mode OFF>

- #14

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you guys are making the problem too complicated

since

[tex]3x^2 + 4x + C = A(x + 1)^2 + B(x + 1) + 7[/tex]

expand [tex]A(x + 1)^2 + B(x + 1) + 7[/tex]

then you have [tex]Ax^2 + 2Ax + A + Bx + B + 7[/tex]

and [tex]3x^2 + 4x + C = Ax^2 + 2Ax + A + Bx + B + 7[/tex]

equate the coeffients of the powers you have

[tex]3 = A[/tex] for [tex]x^2[/tex]

[tex]2A + B = 4[/tex] for [tex]x^1[/tex]

[tex]A + B + 7 = C[/tex] for [tex]x^0[/tex]

solve the system and you got the answer

since

[tex]3x^2 + 4x + C = A(x + 1)^2 + B(x + 1) + 7[/tex]

expand [tex]A(x + 1)^2 + B(x + 1) + 7[/tex]

then you have [tex]Ax^2 + 2Ax + A + Bx + B + 7[/tex]

and [tex]3x^2 + 4x + C = Ax^2 + 2Ax + A + Bx + B + 7[/tex]

equate the coeffients of the powers you have

[tex]3 = A[/tex] for [tex]x^2[/tex]

[tex]2A + B = 4[/tex] for [tex]x^1[/tex]

[tex]A + B + 7 = C[/tex] for [tex]x^0[/tex]

solve the system and you got the answer

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- #16

arildno

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I wasn't exactly pissed off; rather, I felt resignation sneak up on me..

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