C4, binomial expanision

I am doing the paper June 2013(R)
Stuck on 4(b) where it says: using your expansion to estimate an approximate value of (7100)^1/3
For my expansion I got 2-(3/4)x-(9/32)x^2-(45/256)x^3
The Original function is (8-9x)^1/3
If anyone could help it would be good as I don't understand atand what the mark scheme is doing

Reply 1

KaylaB

20

Original post by Dnaap

I am doing the paper June 2013(R)
Stuck on 4(b) where it says: using your expansion to estimate an approximate value of (7100)^1/3
For my expansion I got 2-(3/4)x-(9/32)x^2-(45/256)x^3
The Original function is (8-9x)^1/3
If anyone could help it would be good as I don't understand atand what the mark scheme is doing

This question:

Spoiler

~ Realised I misread one part of the question which made my post wrong ~
:getmecoat:

(edited 8 years ago)

Reply 2

B_9710

18

If you make x=1/10 what do you get inside the cube root?

you get 71/10 so we have

\displaystyle \left (\frac{71}{10} \right )^{1/3} \approx \text{your expansion when x=0.1}

.
Now you need to multiply this by some number to get

\sqrt[3]{7100} \approx \text{your expansion}\times \text{the number you multiplied by on the LHS}

.

Reply 3

Zacken

22

Original post by Dnaap

I am doing the paper June 2013(R)
Stuck on 4(b) where it says: using your expansion to estimate an approximate value of (7100)^1/3
For my expansion I got 2-(3/4)x-(9/32)x^2-(45/256)x^3
The Original function is (8-9x)^1/3
If anyone could help it would be good as I don't understand atand what the mark scheme is doing

7100 = 71 \times 1000

so:

Unparseable latex formula:

\displaystyle[br]\begin{equation*}\sqrt[3]{7100} = \sqrt[3]{71 \times 1000} = \sqrt[3]{71} \times \sqrt[3]{1000} = 10 \times \sqrt[3]{71}\end{equation*}

.

Then, if you let

x = \frac{1}{10}

you get

\displaystyle (8 - 9x)^{1/3} = \left(8 - \frac{9}{10}\right)^{1/3} = \left(\frac{71}{10}\right)^{1/3}

So plugging

x = \frac{1}{10}

into your expansion gets you an approximation for

\left(\frac{71}{10}\right)^{1/3} = 7.1^{1/3}

But

7.1^{1/3} = \frac{\sqrt[3]{7100}}{10} \Rightarrow \sqrt[3]{7100} = 10 \times 7.1^{1/3}

So once you put in

x = 0.1

into your expansion, you approximate

7.1^{1/3}

- you need to multiply it by

10

to get

\sqrt[3]{7100}

.

Reply 4

Dnaap

OP

2

Original post by B_9710

If you make x=1/10 what do you get inside the cube root?

you get 71/10 so we have

\displaystyle \left (\frac{71}{10} \right )^{1/3} \approx \text{your expansion when x=0.1}

.
Now you need to multiply this by some number to get

\sqrt[3]{7100} \approx \text{your expansion}\times \text{the number you multiplied by on the LHS}

.

Why x=1/10 though? How do you know to put 1/10 in

Reply 5

Dnaap

OP

2

Original post by Zacken

7100 = 71 \times 1000

so:

Unparseable latex formula:

\displaystyle[br]\begin{equation*}\sqrt[3]{7100} = \sqrt[3]{71 \times 1000} = \sqrt[3]{71} \times \sqrt[3]{1000} = 10 \times \sqrt[3]{71}\end{equation*}

.

Then, if you let

x = \frac{1}{10}

you get

\displaystyle (8 - 9x)^{1/3} = \left(8 - \frac{9}{10}\right)^{1/3} = \left(\frac{71}{10}\right)^{1/3}

So plugging

x = \frac{1}{10}

into your expansion gets you an approximation for

\left(\frac{71}{10}\right)^{1/3} = 7.1^{1/3}

But

7.1^{1/3} = \frac{\sqrt[3]{7100}}{10} \Rightarrow \sqrt[3]{7100} = 10 \times 7.1^{1/3}

So once you put in

x = 0.1

into your expansion, you approximate

7.1^{1/3}

- you need to multiply it by

10

to get

\sqrt[3]{7100}

.

Thanks very thorough I'm just not sure why you chose to put 1/10 in

Reply 6

Dnaap

OP

2

Original post by Zacken

7100 = 71 \times 1000

so:

Unparseable latex formula:

\displaystyle[br]\begin{equation*}\sqrt[3]{7100} = \sqrt[3]{71 \times 1000} = \sqrt[3]{71} \times \sqrt[3]{1000} = 10 \times \sqrt[3]{71}\end{equation*}

.

Then, if you let

x = \frac{1}{10}

you get

\displaystyle (8 - 9x)^{1/3} = \left(8 - \frac{9}{10}\right)^{1/3} = \left(\frac{71}{10}\right)^{1/3}

So plugging

x = \frac{1}{10}

into your expansion gets you an approximation for

\left(\frac{71}{10}\right)^{1/3} = 7.1^{1/3}

But

7.1^{1/3} = \frac{\sqrt[3]{7100}}{10} \Rightarrow \sqrt[3]{7100} = 10 \times 7.1^{1/3}

So once you put in

x = 0.1

into your expansion, you approximate

7.1^{1/3}

- you need to multiply it by

10

to get

\sqrt[3]{7100}

.

Never mind I got it reading over your working. Thank you a lot!

Reply 7

Zacken

22

Original post by Dnaap

Never mind I got it reading over your working. Thank you a lot!

Glad it helped! :h:

Reply 8

Glavien

11

Original post by Zacken

7100 = 71 \times 1000

so:

Unparseable latex formula:

\displaystyle[br]\begin{equation*}\sqrt[3]{7100} = \sqrt[3]{71 \times 1000} = \sqrt[3]{71} \times \sqrt[3]{1000} = 10 \times \sqrt[3]{71}\end{equation*}

.

Then, if you let