Find the Values of the constants in the following indentities

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  • #1
ASMATHSHELPME
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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:
 
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  • #2
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.
 
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  • #3
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.
 
  • #4
TD: It's [itex]+2x[/itex], not [itex]-2x[/itex] as in your post.
 
  • #5
Right, I'll correct :)
 
  • #6
Bingo, gee that's easy ... but i always think that, then ill try a new one and get confused again.

These books I am 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.
 
  • #7
Seems correct :smile:
 
  • #8
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?
 
  • #9
[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]
 
  • #10
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 each other? (The bottom bits) leaving 2A/ Nothing.
 
  • #11
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.
 
  • #12
Wow that's 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 each other out? Or have i done this basic step wrong in the first place.
 
  • #13
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.
 
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  • #14
So if X^2 is below something(Demoninator i think you call it?) and its multiplyed by itself, It cancels each other 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.
 
  • #15
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]
 
  • #16
ASMATHSHELPME said:
So if X^2 is below something(Demoninator i think you call it?) and its multiplyed by itself, It cancels each other 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:
 
  • #17
ASMATHSHELPME said:
So if X^2 is below something(Demoninator i think you call it?) and its multiplyed by itself, It cancels each other 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.
 

1. What are the constants in an identity?

The constants in an identity are numerical values that remain unchanged throughout the equation. They are usually represented by letters, such as "a" or "c".

2. How do you find the values of the constants in an identity?

To find the values of the constants in an identity, you need to have at least two equations with the same variables. You can then solve the equations simultaneously to determine the values of the constants.

3. Can the values of the constants in an identity change?

No, the values of the constants in an identity cannot change. They are fixed values that remain the same throughout the equation.

4. Are there any strategies for finding the values of the constants in an identity?

Yes, there are several strategies for finding the values of the constants in an identity. These include substitution, elimination, and using matrices.

5. Why is it important to find the values of the constants in an identity?

Finding the values of the constants in an identity is important because it helps us understand the relationship between the variables in the equation. It also allows us to solve for specific values and make predictions based on the equation.

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