This equation is in the formOriginally Posted by SeaN187
and happens to equal so,
substitute
now do the work
and since you can do the same thing to the other side the answer becomes,
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
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
No problem, here's how I would do it.....
the problem starts in the form, , and ends in the form,
so all you need to do is find the value a , , and .
You know, thanks to my work above that so solve for and you get .
Same thing for .
Now solve the entire equation to find , and then put those numbers into the form
Here is the complete work:Originally Posted by SeaN187
extending this, we get:
you can only subtract like terms so,
because each segment has to equal zero, we can split the equation into 5 pieces.
i)
ii)
iii)
iv)
v)
I used equation i) to find "a"
I used equation v) to find "c"
I used equation iv) to find "b"
and all you need to do is substitute those numbers into the answer.
And so we have reached the answer using mathematical steps, proving it.
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