• Jun 10th 2006, 02:58 AM
SeaN187
Pythagoras: a^2+b^2=c^2 which is the same as:
4n^4+8n^3+8n^2+4n+1 = 4n^4+8n^3+8n^2+4n+1
Now use Quadtratic Equations to proove that this is true
(2n^2+2n+1)^2 = (2n^2+2n+1)^2.
It must end as the equation above
Can someone talk me through all the steps plz
• Jun 10th 2006, 05:30 AM
Quick
Quote:

Originally Posted by SeaN187
4n^4+8n^3+8n^2+4n+1 = 4n^4+8n^3+8n^2+4n+1
Now use Quadtratic Equations to proove that this is true
(2n^2+2n+1)^2 = (2n^2+2n+1)^2.
It must end as the equation above
Can someone talk me through all the steps plz

This equation is in the form $\left(ax^2+bx+c\right)^2=\left(ax^2+bx+c\right)^2$
and $\left(ax^2+bx+c\right)^2$ happens to equal $(ax^2)^2+2abx^3+(bx)^2+2acx+2bcx+c^2$ so,

$\left(an^2+bn+c\right)^2=(an^2)^2+2abn^3+(bn)^2+2a cn^2+2bcn+c^2$substitute
$\left(2n^2+4n+1\right)^2$ $=(2n^2)^2+2(2)(2)n^3+(2n)^2+2(2)(1)n^2+2(2)(1)n+1^ 2$ now do the work
$\left(2n^2+4n+1\right)^2=2^2n^{(2\cdot2)}+8n^3+2^2 n^2+4n^2+4n+1$
$\left(2n^2+4n+1\right)^2=4n^4+8n^3+4n^2+4n^2+4n+1$
$\left(2n^2+4n+1\right)^2=4n^4+8n^3+8n^2+4n+1$

and since you can do the same thing to the other side the answer becomes,
$4n^4+8n^3+8n^2+4n+1$ $=4n^4+8n^3+8n^2+4n+1$
• Jun 10th 2006, 05:45 AM
SeaN187
quick yer thnx for tht but the final answer has got to be
(2n^2+2n+1)^2 = (2n^2+2n+1)^2. not
4n^4+8n^3+8n^2+4n+1 = 4n^4+8n^3+8n^2+4n+1

so you've got to work from
4n^4+8n^3+8n^2+4n+1 = 4n^4+8n^3+8n^2+4n+1
to
(2n^2+2n+1)^2 = (2n^2+2n+1)^2

im not trying to make u sound stupid or ought u sound proper smart and i understand what you just did there and thanks for doing it but can you do it like it is above please because i really dont know where to start
• Jun 10th 2006, 05:59 AM
Quick
No problem, here's how I would do it.....
the problem starts in the form, $(ax^2)^2+2abx^3+(bx)^2+2acx+2bcx+c^2$, and ends in the form, $\left(ax^2+bx+c\right)^2$

so all you need to do is find the value a $a$, $b$, and $c$.
You know, thanks to my work above that $(ax^2)^2=4x^4$ so solve for $a$ and you get $a=2$.
Same thing for $c$.
$c^2=1$
$c=1$

Now solve the entire equation to find $b$, and then put those numbers into the form $\left(ax^2+bx+c\right)^2$
• Jun 10th 2006, 06:01 AM
SeaN187
isn't that showing that the formula is true not prooving it?
• Jun 10th 2006, 06:20 AM
Quick
Quote:

Originally Posted by SeaN187
isn't that showing that the formula is true not prooving it?

I have found the answer mathematically, instead of guessing, and that proves it. I could show you the complete work I did if you want.
• Jun 10th 2006, 06:58 AM
SeaN187
yeh n u show me all the work u did plz
n can i ask u a question why does everyone do their equations in tht big writting javascript thing
• Jun 10th 2006, 07:06 AM
CaptainBlack
Quote:

Originally Posted by SeaN187
Pythagoras: a^2+b^2=c^2 which is the same as:
4n^4+8n^3+8n^2+4n+1 = 4n^4+8n^3+8n^2+4n+1
Now use Quadtratic Equations to proove that this is true
(2n^2+2n+1)^2 = (2n^2+2n+1)^2.
It must end as the equation above
Can someone talk me through all the steps plz

Since both the RHS and LHS are equal this is trivial true:

$(2n^2+2n+1)^2 = (2n^2+2n+1)^2$

RonL
• Jun 10th 2006, 07:06 AM
Quick
Quote:

Originally Posted by SeaN187
yeh n u show me all the work u did plz
n can i ask u a question why does everyone do their equations in tht big writting javascript thing

Here is the complete work:

$4n^4+8n^3+8n^2+4n+1$ $=a^2n^4+2abn^3+b^2n^2+2acn^2+2bcn+c^2$ extending this, we get:
$4n^4+8n^3+8n^2+4n+1$ $-\left(a^2n^2+2abn^3+b^2n^2+2a cn^2+2bcn+c^2\right)=0$ you can only subtract like terms so,
$4n^4-a^2n^4+8n^3-2abn^3$ $+8n^2-(b^2n^2+2acn^2)+4n-2bcn+1-c^2=0$ because each segment has to equal zero, we can split the equation into 5 pieces.
i) $4n^4-a^2n^4=0$
ii) $8n^3-2abn^3=0$
iii) $8n^2-(b^2n^2+2acn^2)=0$
iv) $4n-2bcn=0$
v) $1-c^2=0$

I used equation i) to find "a"
$4n^4-a^2n^4=0$
$4n^4=a^2n^4$
$4=a^2$
$\sqrt{4}=a$
$2=a$

I used equation v) to find "c"
$1-c^2=0$
$1=c^2$
$\sqrt{1}=c$
$1=c$

I used equation iv) to find "b"
$4n-2bcn=0$
$4n=2bcn$
$4n=2b(1)n$
$4n=2bn$
$2n=bn$
$2=b$

and all you need to do is substitute those numbers into the answer.
$\left(an^2+bn+c\right)^2$
$\left(2n^2+2n+1\right)^2$

And so we have reached the answer using mathematical steps, proving it.

We use the LaTex code because it is much easier to understand than other normal text.
• Jun 10th 2006, 08:06 AM
SeaN187
yer thanks alot a totally understand now
• Jun 10th 2006, 08:11 AM
SeaN187
this isnt using quadratic equations though is it? i think he said i have to use quadratic equations
• Jun 10th 2006, 08:21 AM
CaptainBlack
Quote:

Originally Posted by SeaN187
Pythagoras: a^2+b^2=c^2 which is the same as:
4n^4+8n^3+8n^2+4n+1 = 4n^4+8n^3+8n^2+4n+1
Now use Quadtratic Equations to proove that this is true
(2n^2+2n+1)^2 = (2n^2+2n+1)^2.
It must end as the equation above
Can someone talk me through all the steps plz

Divide both sideds of:

$4n^4+8n^3+8n^2+4n+1=4n^4+8n^3+8n^2+4n+1$

by:

$2n^2+2n+1$,

this gives:

$2n^2+2n+1=2n^2+2n+1$,

now square:

$(2n^2+2n+1)^2=(2n^2+2n+1)^2$.

RonL :D

PS check the question, you appear to be asking us to prove $x=x$.
• Jun 10th 2006, 10:15 AM
SeaN187
cheers that is great captain black
thanks to both captain black and quick luv yaxxx
o n captain black im proving a^2+b^2=c^2
• Jun 10th 2006, 07:18 PM
ThePerfectHacker
Quote:

Originally Posted by SeaN187
isn't that showing that the formula is true not prooving it?

I really do not see what the problem is.

I understand what you are saying that is was not mathematically derived but as CaptainBlack said because if expanded it is equal the formula is proved.