# Binomial Expansion Help?

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#1

Hi I'm confused with part (b) to this question which is from the replacement June 2013 C4 paper. My expansion to part (a) is 2- 3/4x - 9/32x^2 - 45/256x^3

So for part (b) I did:

( 8 - 9x )^1/3 = ( 7100 )^1/3

8 - 9x = 7100

x = -788

And when I substituted this into the expansion, I got a really weird answer, which was wrong. The mark scheme says x=0.1

How do you go about answering this and why is my method wrong? Thank you
1
7 years ago
#2
(Original post by Varnie0x)

x = -788
x has to be less than 8/9 otherwise the expansion is nonsense
1
7 years ago
#3
You mean the magnitude of x, |x| < 8/9 as stated in the question.

(Original post by TenOfThem)
x has to be less than 8/9 otherwise the expansion is nonsense
0
7 years ago
#4
According to the mark scheme:

This is obviously impossible. Therefore I propose that the mark scheme is incorrect.
1
7 years ago
#5
Yes, I agree there is a typo. However I think the value of x is correct. The error is in the cube root. It should read (7.1)^1/3 not 71.

(Original post by samsama)
According to the mark scheme:

This is obviously impossible. Therefore I propose that the mark scheme is incorrect.
1
7 years ago
#6
(Original post by Varnie0x)

Hi I'm confused with part (b) to this question which is from the replacement June 2013 C4 paper. My expansion to part (a) is 2- 3/4x - 9/32x^2 - 45/256x^3

So for part (b) I did:

( 8 - 9x )^1/3 = ( 7100 )^1/3

8 - 9x = 7100

x = -788

And when I substituted this into the expansion, I got a really weird answer, which was wrong. The mark scheme says x=0.1

How do you go about answering this and why is my method wrong? Thank you
7100=8000-900=1000(8-0.9)
Does this help?
0
#7
(Original post by WishingChaff)
Yes, I agree there is a typo. However I think the value of x is correct. The error is in the cube root. It should read (7.1)^1/3 not 71.
So it's 7.1 as the modulus of x<8/9?
0
7 years ago
#8
(Original post by Varnie0x)
So it's 7.1 as the modulus of x<8/9?
yes
0
7 years ago
#9
(Original post by Varnie0x)
So it's 7.1 as the modulus of x<8/9?
To use your binomial expansion, you need a value of x with |x| < 8/9. Obviously you can't achieve this starting from 7100 itself, but 7100 = 1000 x 7.1 and you know what the cube root of 1000 is, so you work with 7.1 which gives you an x value of 0.1 i.e. small enough for the binomial expansion to be valid.
1
#10
(Original post by davros)
To use your binomial expansion, you need a value of x with |x| < 8/9. Obviously you can't achieve this starting from 7100 itself, but 7100 = 1000 x 7.1 and you know what the cube root of 1000 is, so you work with 7.1 which gives you an x value of 0.1 i.e. small enough for the binomial expansion to be valid.
Oh I see, thank you
0
3 years ago
#11
(Original post by davros)
To use your binomial expansion, you need a value of x with |x| < 8/9. Obviously you can't achieve this starting from 7100 itself, but 7100 = 1000 x 7.1 and you know what the cube root of 1000 is, so you work with 7.1 which gives you an x value of 0.1 i.e. small enough for the binomial expansion to be valid.
I've just been stuck on this question (my C4 exam is tomorrow lol) and this explanation really helped, thanks!!!!!
0
3 years ago
#12
Is this A2?
0
3 years ago
#13
(Original post by katerh)
I've just been stuck on this question (my C4 exam is tomorrow lol) and this explanation really helped, thanks!!!!!
lol same gl tomorrow
0
3 years ago
#14
(Original post by gerib17)
Is this A2?
Yes.
0
1 year ago
#15
Does anyone know how to do part a? I know how to expand it, I'm just thrown off by the cube root. Thanks
0
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