help needed (maths)

[sticksaint]

My daughter has a question:

 

 

Prove that (n+5)-(n+3)squared = 4(n+4).

 

 

Any help appreciated .Thanks

[Weston Super Saint]

http://www.wolframalpha.com/input/?i=Prove+that+%28n%2B5%29-%28n%2B3%29squared+%3D+4%28n%2B4%29.

[GenevaSaint]

I thought you were the intelligent one in the family stick.

[Whitey Grandad]

My daughter has a question:

 

 

Prove that (n+5)-(n+3)squared = 4(n+4).

 

 

Any help appreciated .Thanks

I don't think it does, as it is written. Are you sure you've got it right?

[SuperMikey]

PM Ludwig, he'll know.

[sticksaint]

I don't think it does, as it is written. Are you sure you've got it right?

 

 

Yes that is the question ,i`ve just checked. But i`m stuck i can`t prove it

[sticksaint]

Thanks guys.Just worked it out -it was a misprint.

 

Question should read prove (n+5) squared - (n+3) squared = 4(n+4)

 

Special thanks to whitey grandad for making me realise it was a misprint!

[Badger]

First draft.

 

n=2.

 

(2+5)sq -(2+3)sq = 24 = 4(2+4)

 

(49)-(25)=24= 4x6

Edited 3 June, 2010 by Badger

[Draino76]

Cool. I can also confirm the original question wasted 5 minutes of my life with no answer.

[Joensuu]

Thanks guys.Just worked it out -it was a misprint.

 

Question should read prove (n+5) squared - (n+3) squared = 4(n+4)

 

Special thanks to whitey grandad for making me realise it was a misprint!

 

Are you sure thats the question? As I think that only works out to be true when n is -4

 

Edit: think I'm wrong: (n+5)squared -(n+3)squared = 4(n+4) is true...

Edited 3 June, 2010 by Joensuu

[Draino76]

n = 0

[Hatch]

n could equal anything I reckon.

[bolo]

(N+5)(N+5) - (N+3)(N+3) =

Nsqr +10N+25 - (Nsqr +6N+9) =

4N + 16 = 4 (N+4)

 

 

God i have too much time on my hands....

[Badger]

n could equal anything I reckon.

 

think you might be right.

 

Remember though that the question is to prove the rule.

 

Not find out the value of n.

[RedAndWhite91]

Ah I remember learning this in school. I use the term 'learning' loosely. I still don't have bloody clue what they are or how to do them and I haven't had to use it since I left school. Me 1-0 School.

[Colinjb]

My daughter has a question:

 

 

Prove that (n+5)-(n+3)squared = 4(n+4).

 

 

Any help appreciated .Thanks

 

( (n+5) - (n+3) )^2 = Squaring something means times it by itself therefore...

( (n+5)*(n+5) ) - ( (n+3)*(n+3) ) = Now multiply out

(n^2 + 5n + 5n + 25) - ( n^2 + 3n + 3n + 9 ) = Simplify

(n^2 + 10n + 25) - ( n^2 + 6n + 9) = Subtract the values that match from each other

4n + 16 = Check what divides through both values

4(n+4)

Edited 3 June, 2010 by Colinjb

[Hatch]

think you might be right.

 

Remember though that the question is to prove the rule.

 

Not find out the value of n.

 

yep, I got it down to 4n + 16 = 4n + 16.

[thesaint sfc]

edit

Edited 3 June, 2010 by thesaint sfc
Oh...we're in The Lounge

[Marsdinho]

nerd alert

[Glasgow_Saint]

N = Na, na, na, na, na, na, na, na, na, na, na baby give it up, give it up

[Colinjb]

N = Na, na, na, na, na, na, na, na, na, na, na baby give it up, give it up

 

Ra-si-ak, Ra-siak? Gregorz Rasiak?

[Ludwig]

think you might be right.

 

Remember though that the question is to prove the rule.

 

Not find out the value of n.

 

Something vaguely interesting involving a different method to the one used before would be to use inductive reasoning to show it is true for all n (though that wouldn't be expected at what appears to be GCSE level maths (it's not even in the A Level Maths syllabus, which is a little ridiculous)).

 

Essentially, if you show that the statement holds true for n=1, then demonstrate that if we accept that it holds for n=k then that implies that it would also hold for n=k+1, then it holds for all n. Because, from the reasoning followed, as it holds for 1, and holding for any 'k' implies it holds for 'k+1', it must hold for 2,3,4,5..... like a domino effect. Though, in this problem, you'd end up multiplying out to show that p(k) => p(k+1), so it's somewhat unnecessary in this situation, but interesting (and more rigorous) nonetheless.

[bridge too far]

+1

[sfcuk fan]

wooooooosh , reet over my head

[Al de Man]

Something vaguely interesting involving a different method to the one used before would be to use inductive reasoning to show it is true for all n (though that wouldn't be expected at what appears to be GCSE level maths (it's not even in the A Level Maths syllabus, which is a little ridiculous)).

 

Essentially, if you show that the statement holds true for n=1, then demonstrate that if we accept that it holds for n=k then that implies that it would also hold for n=k+1, then it holds for all n. Because, from the reasoning followed, as it holds for 1, and holding for any 'k' implies it holds for 'k+1', it must hold for 2,3,4,5..... like a domino effect. Though, in this problem, you'd end up multiplying out to show that p(k) => p(k+1), so it's somewhat unnecessary in this situation, but interesting (and more rigorous) nonetheless.

 

We covered Proof by Induction as the very first subject for A-level Pure Maths and it made we want to jack it in.

[StuRomseySaint]

Why use letters? Isn't that what numbers are for? :confused:

[Colinjb]

Why use letters? Isn't that what numbers are for? :confused:

:smt069:rolleyes:

[Picard]

This is the difference of two squares rule:

 

a² - b² = (a+b)*(a-b)

[Glasgow_Saint]

Ra-si-ak, Ra-siak? Gregorz Rasiak?

 

Nope....

 

Na, na, na, na, na, na, na, na, na, na, na David Connolly, Connolly, David Connolly Na, na, na, na, na, na, na, na, na, na, na!

 

[Dark Munster]

[sticksaint]

 

 

My daughter has her Gcse maths exam tomorrow.Thanks dark munster for providing some tuition :-)

[Deano6]

( (n+5) - (n+3) )^2 = Squaring something means times it by itself therefore...

( (n+5)*(n+5) ) - ( (n+3)*(n+3) ) = Now multiply out

(n^2 + 5n + 5n + 25) - ( n^2 + 3n + 3n + 9 ) = Simplify

(n^2 + 10n + 25) - ( n^2 + 6n + 9) = Subtract the values that match from each other

4n + 16 = Check what divides through both values

4(n+4)

 

FAIL!

 

( (n+5) - (n+3) )^2

 

does not equal

 

( (n+5)*(n+5) ) - ( (n+3)*(n+3) )

 

it equals

 

(n+5)*(n+5) ) - (n+3)*(n+3) - 2 * (n+5) (n+3)

 

 

That wasn't the question.

 

 

Something vaguely interesting involving a different method to the one used before would be to use inductive reasoning to show it is true for all n (though that wouldn't be expected at what appears to be GCSE level maths (it's not even in the A Level Maths syllabus, which is a little ridiculous)).

 

Essentially, if you show that the statement holds true for n=1, then demonstrate that if we accept that it holds for n=k then that implies that it would also hold for n=k+1, then it holds for all n. Because, from the reasoning followed, as it holds for 1, and holding for any 'k' implies it holds for 'k+1', it must hold for 2,3,4,5..... like a domino effect. Though, in this problem, you'd end up multiplying out to show that p(k) => p(k+1), so it's somewhat unnecessary in this situation, but interesting (and more rigorous) nonetheless.

 

Not relevant here - you are just showing off. Also, this wouldn't be a satisfactory answer as you have only proved it for positive integers, not all real numbers, which would be impossible by induction.

 

(N+5)(N+5) - (N+3)(N+3) =

Nsqr +10N+25 - (Nsqr +6N+9) =

4N + 16 = 4 (N+4)

 

 

God i have too much time on my hands....

 

Correct

[Ponty]

[Ludwig]

Not relevant here - you are just showing off. Also, this wouldn't be a satisfactory answer as you have only proved it for positive integers, not all real numbers, which would be impossible by induction.

 

fml