View Full Version : Indentities
ASMATHSHELPME
Aug30-05, 07:41 AM
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:
So, you start with
A\left( {x^2 - 1} \right) + B\left( {x - 1} \right) + C = \left( {3x - 1} \right)\left( {x + 1} \right)
You worked out both sides
Ax^2 - A + Bx - B + C = 3x^2 + 3x - x - 1
Now, before you continue, rearrange both sides so you group per power of x
Ax^2 + Bx - A - B + C = 3x^2 + 2x - 1
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.
iNCREDiBLE
Aug30-05, 07:50 AM
Just a small error. You forgot to simplify 3x-x to 2x.
A(x^2-1) + B(x-1) + C = (3x-1)(x+1)
<=>
Ax^2-A+Bx-B+C = 3x^2+3x-x-1 = 3x^2+2x-1
<=>
Ax^2+Bx-A-B+C = 3x^2+2x-1
Now try again.
iNCREDiBLE
Aug30-05, 07:53 AM
TD: It's +2x, not -2x as in your post.
ASMATHSHELPME
Aug30-05, 08:05 AM
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.
ASMATHSHELPME
Aug30-05, 08:19 AM
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????
iNCREDiBLE
Aug30-05, 08:26 AM
(a-b)(a-b) = (a-b)^2 = a^2 - 2ab + b^2
So -2x^2\frac{A}{x^2}=-2A[/tex] not [itex]2A/X^8
ASMATHSHELPME
Aug30-05, 08:32 AM
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.
iNCREDiBLE
Aug30-05, 08:43 AM
I assume you know that (a-b)(a-b) = (a-b)^2 = a^2 - 2ab + b^2 (*)
So:
(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}
What I'm saying is that:
-2x^2\frac{A}{x^2} = \frac{-2Ax^2}{x^2} = -2A
since the x^2 in the numerator and denominator cancels out.
ASMATHSHELPME
Aug30-05, 08:51 AM
Wow thats confusing!
So can you quickly type out (X^2 - A/x^2)^2 Please
Because i get:
+x^4 - A/x^4 - A/x^4 + A^2/x^4
Just want to know how X^4 cancel eachother out? Or have i done this basic step wrong in the first place.
iNCREDiBLE
Aug30-05, 08:57 AM
Yes, you seem to believe that
x^2\frac{A}{x^2} = \frac{A}{x^4}
which is not true.
In fact x^2\frac{A}{x^2} = A
So
(x^2-A/x^2)(x^2-A/x^2) = x^4 - A - A + A^2/X^4
[Edit]: Missed one +-sign.
ASMATHSHELPME
Aug30-05, 09:05 AM
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 X^2 [Multiplied by ] A/X^2 Cancels X^2 leaving A/ Nothing.
iNCREDiBLE
Aug30-05, 11:37 AM
That's right. Just take a look at this trivial example:
1 = \frac{1}{2} + \frac{1}{2} = (definition) = 2*\frac{1}{2} = 1
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 X^2 [Multiplied by ] A/X^2 Cancels X^2 leaving A/ Nothing.
A little bit of terminology:
- we call \frac{A}{B} a fraction
- A is the nominator
- B is the denominator
Perhaps that makes it easier to follow :smile:
HallsofIvy
Aug30-05, 12:04 PM
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 X^2 [Multiplied by ] A/X^2 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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