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# Validity involving Binomial C4 Watch

1. I do not get how to do validity, once we expand the expanssion, how do we get the validity ?
2. (Original post by otrivine)
I do not get how to do validity, once we expand the expanssion, how do we get the validity ?
When written in the form a(1+bx)^c, the magnitude of bx must be smaller than 1.
e.g. for (4+8x)^0.5 = 2(1+2x)^0.5:
-1<2x<1, so -0.5<x<0.5
3. (Original post by TenOfThem)
The magnitude of the bracket has to be less than 1
thanks for example if the binomial was

4/(2+3x)

I expanded and got
2-3x + 9/2x2 - 27/4 x3

so then what do I do for validity? do I solve 2+3x=0 to get -2/3 so |2/3|<1 ?
4. (Original post by otrivine)
thanks for example if the binomial was

4/(2+3x)

I expanded and got
2-3x + 9/2x2 - 27/4 x3

so then what do I do for validity? do I solve 2+3x=0 to get -2/3 so |2/3|<1 ?
You don't need to expand the bracket for validity.
Just look at the bracket bit.
(2+3x)^-1
now take the two outside the bracket, by dividing everything within the bracket by 2:
(2^-1)(1+1.5x)^-1
Now your bracket is in a more useable form. The magnitude of 1.5x must be smaller than 1:
-1<1.5x<1
therefore:
-2/3<x<2/3
5. (Original post by dragonkeeper999)
You don't need to expand the bracket for validity.
Just look at the bracket bit.
(2+3x)^-1
now take the two outside the bracket, by dividing everything within the bracket by 2:
(2^-1)(1+1.5x)^-1
Now your bracket is in a more useable form. The magnitude of 1.5x must be smaller than 1:
-1<1.5x<1
therefore:
-2/3<x<2/3

Cause in the book they express lie 1<|2/3| ?
6. (Original post by TenOfThem)
The magnitude of the bracket has to be less than 1
The magnitude of the bracket or the magnitude of x?
(1.01)^0.5 has a magnitude greater than one but will converge if (1+x)^0.5 is expanded and x = 0.01

0.01 is sufficiently small.
7. (Original post by otrivine)

Cause in the book they express lie x<|2/3| ?
That is also a correct way of representing -2/3<x<2/3 (assuming you meant x rather than 1 above!). I don't know how to get my laptop to write the modulus symbols (|) though.
EDIT: actually, no. I think the modulus symbols should be around the x not the 2/3:
|x|<2/3
8. (Original post by dragonkeeper999)
That is also a correct way of representing -2/3<x<2/3 (assuming you meant x rather than 1 above!). I don't know how to get my laptop to write the modulus symbols (|) though.
EDIT: actually, no. I think the modulus symbols should be around the x not the 2/3:
|x|<2/3
Perfect understood

are you doing c4 in june, because I have a question on mixed execise 3D question 11)a) partial fraction, I THINk the book made a mistake cause it is improper and they did not do the long division /
9. (Original post by otrivine)
Perfect understood

are you doing c4 in june, because I have a question on mixed execise 3D question 11)a) partial fraction, I THINk the book made a mistake cause it is improper and they did not do the long division /
I'm doing C4 in June, yes OCR MEI though, what exam board are you on? Post the question up with your working for me to have a look at. And there's nothing wrong with improper fractions...
10. (Original post by dragonkeeper999)
I'm doing C4 in June, yes OCR MEI though, what exam board are you on? Post the question up with your working for me to have a look at. And there's nothing wrong with improper fractions...
I am doing edexcel No cause the book made a mistake I think they should have used long division first and then apply the partial fraction method

Express (9x2+26x+20)/(1+x)(2+x)2
as partial fractions
11. (Original post by otrivine)
I am doing edexcel No cause the book made a mistake I think they should have used long division first and then apply the partial fraction method

Express (9x2+26x+20)/(1+x)(2+x)2
as partial fractions
Long division?!!!
Use the A/1+x + B/2+x + C/(2+x)2 method. Show some working so I can suggest where you may be going wrong...
12. (Original post by otrivine)
I am doing edexcel No cause the book made a mistake I think they should have used long division first and then apply the partial fraction method

Express (9x2+26x+20)/(1+x)(2+x)2
as partial fractions
No tricks here. imo, it looks like you can go ahead and use your normal methods here. If I say A + B + C, does that help?
13. (Original post by dragonkeeper999)
Long division?!!!
Use the A/1+x + B/2+x + C/(2+x)2 method. Show some working so I can suggest where you may be going wrong...
for example ok sub 5 into x and you get 125/98 which means that this is improper, agree or not so far
14. (Original post by otrivine)
for example ok sub 5 into x and you get 125/98 which means that this is improper, agree or not so far
What?!

You have to multiply through by the denominator.
15. (Original post by otrivine)
for example ok sub 5 into x and you get 125/98 which means that this is improper, agree or not so far
It doesn't matter if it is improper...
If it helps, my answer is:
3/(1+x) + 6/(2+x) - 4/ (2+x)2 Does this agree with your book's answer or with yours?
Post some more working so we can understand what you are doing.
16. (Original post by Hype en Ecosse)
No tricks here. imo, it looks like you can go ahead and use your normal methods here. If I say A + B + C, does that help?
Hi, how are you

Ok, for example , I posted above , if you sub x=5 right into the equation you get improper fraction of 125/98 , agree?
17. (Original post by dragonkeeper999)
It doesn't matter if it is improper...
If it helps, my answer is:
3/(1+x) + 6/(2+x) - 4/ (2+x)2 Does this agree with your book's answer or with yours?
Yes, that what the book says, but some strange reason in my book it mentions if the value on the top (denomintaor) is more than the numerator you use long division

for example this one here they use long division

(3x2-3x-2)/(x-1)(x-2)

do you see where I am getting confused
18. (Original post by otrivine)
for example ok sub 5 into x and you get 125/98 which means that this is improper, agree or not so far
When it comes to algebraic expressions, it counts as improper if the highest power of the numerator is greater than or equal to the highest power of the denominator. You don't have to divide here because the power of the numerator (2) is less than the power of the denominator (3).

In the post above, that one's improper because the highest powers are equal. In the original, the highest power of the denominator is larger. So it's proper.
19. (Original post by otrivine)
Hi, how are you

Ok, for example , I posted above , if you sub x=5 right into the equation you get improper fraction of 125/98 , agree?
You're getting this all mixed up.

Now multiply through by the denominator on the LHS.

This is the standard method. It's proper because there's a cubic in the denominator and a quadratic in the numerator.
20. (Original post by otrivine)
Yes, that what the book says, but some strange reason in my book it mentions if the value on the top (denomintaor) is more than the numerator you use long division

for example this one here they use long division

(3x2-3x-2)/(x-1)(x-2)

do you see where I am getting confused
Because that one IS an improper algebraic fraction.
Someone above explains it a bit better...
Anyway, going back to the actual question, can you please post some proper working?

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