Find the Values of the constants in the following indentities

Main Question or Discussion Point

Find the Values of the constants in the following indentities.

A(X^2 -1) + B(X - 1) + C = (3x -1)(x +1)

AX^2 - A + Bx - B + C = 3x^2 + 3x -x -1

Ok, Is this correct so far?

AX^2 = 3x^2 :. A = 3
Is correct ?

Bx = 3x = b = 3 ?
C = -1 ?

I'm sure this is wrong, I just need help explaining why and correcting my errors. It seems i need to simplify it more before comparing the co-efficients.

Thanks in advanced guys,
Probably all find this easy :blushing:
 

Answers and Replies

TD
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So, you start with

[tex]A\left( {x^2 - 1} \right) + B\left( {x - 1} \right) + C = \left( {3x - 1} \right)\left( {x + 1} \right)[/tex]

You worked out both sides

[tex]Ax^2 - A + Bx - B + C = 3x^2 + 3x - x - 1[/tex]

Now, before you continue, rearrange both sides so you group per power of x

[tex]Ax^2 + Bx - A - B + C = 3x^2 + 2x - 1[/tex]

Now compare the coefficients of the equal powers of x of both sides. Remember that the all the constants are the coefficients of x^0.
 
Last edited:
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Just a small error. You forgot to simplify [itex]3x-x[/itex] to [itex]2x[/itex].

[tex]A(x^2-1) + B(x-1) + C = (3x-1)(x+1)[/tex]
<=>
[tex]Ax^2-A+Bx-B+C = 3x^2+3x-x-1 = 3x^2+2x-1 [/tex]
<=>
[tex]Ax^2+Bx-A-B+C = 3x^2+2x-1 [/tex]

Now try again.
 
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TD: It's [itex]+2x[/itex], not [itex]-2x[/itex] as in your post.
 
TD
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Right, I'll correct :)
 
Bingo, gee thats easy ... but i always think that, then ill try a new one and get confused again.

These books im learning from arn't to clear.

Thanks guys,
Ax^2 + Bx - A - B +C = 3x^2 + 2x - 1
A = 3
B = 2
-A -B + C = -1
-3 -2 + C = - 1
C = 4
-5 + 4 = - 1

Excellent.
 
TD
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Seems correct :smile:
 
One more question, a new identity but basically expanding with fractions.

(X^2 - A/X^2) (X^2 - A/X^2)

= X^4 - A/X^4 - A/X^4 + A^2/X^4

X^4 - 2A/x^8 + A^2/X^4 ???

But is the 2A/X^8 Correct? Sure its wrong????
 
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[itex](a-b)(a-b) = (a-b)^2 = a^2 - 2ab + b^2 [/itex]

So [itex]-2x^2\frac{A}{x^2}=-2A[/tex] not [itex]2A/X^8[/itex]
 
How do you make it look like that?

My layouits are unclear so i think you've given the wrong explanation(My fault).

Wondering how its 2A/Nothing

Rather than
2A/x^8

From -A/x^4 -A/X^4

SOmehow they cancel eachother? (The bottom bits) leaving 2A/ Nothing.
 
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I assume you know that [itex](a-b)(a-b) = (a-b)^2 = a^2 - 2ab + b^2[/itex] (*)

So:
[tex](x^2 - \frac{A}{x^2}) (x^2 - \frac{A}{x^2}) = (x^2 - \frac{A}{x^2})^2 = (*) = (x^2)^2 -2x^2\frac{A}{x^2} + (\frac{A}{x^2})^2 = x^4 - 2A + \frac{A^2}{x^4}[/tex]

What I'm saying is that:
[tex]-2x^2\frac{A}{x^2} = \frac{-2Ax^2}{x^2} = -2A[/tex]
since the [itex]x^2[/itex] in the numerator and denominator cancels out.
 
Wow thats confusing!

So can you quickly type out[itex] (X^2 - A/x^2)^2 [/itex] Please

Because i get:

[itex]+x^4 - A/x^4 - A/x^4 + A^2/x^4[/itex]
Just want to know how X^4 cancel eachother out? Or have i done this basic step wrong in the first place.
 
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Yes, you seem to believe that
[tex]x^2\frac{A}{x^2} = \frac{A}{x^4}[/tex]
which is not true.

In fact [itex]x^2\frac{A}{x^2} = A[/itex]

So
[tex](x^2-A/x^2)(x^2-A/x^2) = x^4 - A - A + A^2/X^4[/tex]


[Edit]: Missed one +-sign.
 
Last edited:
So if X^2 is below something(Demoninator i think you call it?) and its multiplyed by itself, It cancels eachother out?

Just trying to get my head around this, Quite confused lol.

Basically [itex]X^2 [Multiplied by ] A/X^2[/itex] Cancels X^2 leaving A/ Nothing.
 
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That's right. Just take a look at this trivial example:

[tex]1 = \frac{1}{2} + \frac{1}{2} = (definition) = 2*\frac{1}{2} = 1[/tex]
 
TD
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ASMATHSHELPME said:
So if X^2 is below something(Demoninator i think you call it?) and its multiplyed by itself, It cancels eachother out?

Just trying to get my head around this, Quite confused lol.

Basically [itex]X^2 [Multiplied by ] A/X^2[/itex] Cancels X^2 leaving A/ Nothing.
A little bit of terminology:

- we call [itex]\frac{A}{B}[/itex] a fraction
- A is the nominator
- B is the denominator

Perhaps that makes it easier to follow :smile:
 
HallsofIvy
Science Advisor
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ASMATHSHELPME said:
So if X^2 is below something(Demoninator i think you call it?) and its multiplyed by itself, It cancels eachother out?

Just trying to get my head around this, Quite confused lol.

Basically [itex]X^2 [Multiplied by ] A/X^2[/itex] Cancels X^2 leaving A/ Nothing.
Well, not A over Nothing! That would be A/0 which makes no sense.
The cancelling leaves 1. x2(A/x2= A/1= A.
 

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