• Jun 10th 2006, 01:58 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, 04:30 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 $\displaystyle \left(ax^2+bx+c\right)^2=\left(ax^2+bx+c\right)^2$
and $\displaystyle \left(ax^2+bx+c\right)^2$ happens to equal $\displaystyle (ax^2)^2+2abx^3+(bx)^2+2acx+2bcx+c^2$ so,

$\displaystyle \left(an^2+bn+c\right)^2=(an^2)^2+2abn^3+(bn)^2+2a cn^2+2bcn+c^2$substitute
$\displaystyle \left(2n^2+4n+1\right)^2$$\displaystyle =(2n^2)^2+2(2)(2)n^3+(2n)^2+2(2)(1)n^2+2(2)(1)n+1^ 2 now do the work \displaystyle \left(2n^2+4n+1\right)^2=2^2n^{(2\cdot2)}+8n^3+2^2 n^2+4n^2+4n+1 \displaystyle \left(2n^2+4n+1\right)^2=4n^4+8n^3+4n^2+4n^2+4n+1 \displaystyle \left(2n^2+4n+1\right)^2=4n^4+8n^3+8n^2+4n+1 and since you can do the same thing to the other side the answer becomes, \displaystyle 4n^4+8n^3+8n^2+4n+1$$\displaystyle =4n^4+8n^3+8n^2+4n+1$
• Jun 10th 2006, 04:45 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, 04:59 AM
Quick
No problem, here's how I would do it.....
the problem starts in the form, $\displaystyle (ax^2)^2+2abx^3+(bx)^2+2acx+2bcx+c^2$, and ends in the form, $\displaystyle \left(ax^2+bx+c\right)^2$

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

Now solve the entire equation to find $\displaystyle b$, and then put those numbers into the form $\displaystyle \left(ax^2+bx+c\right)^2$
• Jun 10th 2006, 05:01 AM
SeaN187
isn't that showing that the formula is true not prooving it?
• Jun 10th 2006, 05:20 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, 05:58 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, 06:06 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:

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

RonL
• Jun 10th 2006, 06:06 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:

$\displaystyle 4n^4+8n^3+8n^2+4n+1$$\displaystyle =a^2n^4+2abn^3+b^2n^2+2acn^2+2bcn+c^2 extending this, we get: \displaystyle 4n^4+8n^3+8n^2+4n+1$$\displaystyle -\left(a^2n^2+2abn^3+b^2n^2+2a cn^2+2bcn+c^2\right)=0$ you can only subtract like terms so,
$\displaystyle 4n^4-a^2n^4+8n^3-2abn^3$$\displaystyle +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)$\displaystyle 4n^4-a^2n^4=0$
ii)$\displaystyle 8n^3-2abn^3=0$
iii)$\displaystyle 8n^2-(b^2n^2+2acn^2)=0$
iv)$\displaystyle 4n-2bcn=0$
v)$\displaystyle 1-c^2=0$

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

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

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

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

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, 07:06 AM
SeaN187
yer thanks alot a totally understand now
• Jun 10th 2006, 07:11 AM
SeaN187
this isnt using quadratic equations though is it? i think he said i have to use quadratic equations
• Jun 10th 2006, 07:21 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:

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

by:

$\displaystyle 2n^2+2n+1$,

this gives:

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

now square:

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

RonL :D

PS check the question, you appear to be asking us to prove $\displaystyle x=x$.
• Jun 10th 2006, 09:15 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, 06:18 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.