Quadratic Equations

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June 10th 2006, 01:58 AM
SeaN187

Quadratic Equations

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
June 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 $\left(ax^2+bx+c\right)^2=\left(ax^2+bx+c\right)^2$
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,
$4n^4+8n^3+8n^2+4n+1$ $=4n^4+8n^3+8n^2+4n+1$
June 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
June 10th 2006, 04:59 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 $\left(ax^2+bx+c\right)^2$
June 10th 2006, 05:01 AM
SeaN187

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

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

RonL
June 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:

$4n^4+8n^3+8n^2+4n+1$ $=a^2n^4+2abn^3+b^2n^2+2acn^2+2bcn+c^2$ 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.
June 10th 2006, 07:06 AM
SeaN187

yer thanks alot a totally understand now
June 10th 2006, 07:11 AM
SeaN187

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

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

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 $x=x$ .
June 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
June 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.