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

2. 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$

3. 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

4. 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 $\displaystyle \left(ax^2+bx+c\right)^2$

5. isn't that showing that the formula is true not prooving it?

6. 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.

7. 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

8. 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

9. 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.

10. yer thanks alot a totally understand now

11. this isnt using quadratic equations though is it? i think he said i have to use quadratic equations

12. 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

PS check the question, you appear to be asking us to prove $\displaystyle x=x$.

13. 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

14. 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.