are theseequations equal or not?

ryan750

4(x-4)^2 + (2y -5)^2 = 25

and

(x-4)^2 + (y - 2.5)^2 = 25

and if so - can any 1 explain pls?

Reply 1

El Stevo

4(x-4)^2 + (2y -5)^2 = 25
(x-4)^2 + (y - 2.5)^2 = 25

no... see expandage...

4(x^2) - 32x + 64 + 4(y^2) - 20y + 25 = 25
4(x^2) - 32x + 4(y^2) - 20y = -64

(x^2) - 8x + 16 + (y^2) - 5y + 6.25 = 25
(x^2) - 8x + (y^2) - 5y = 2.75

4(x^2) - 32x + 4(y^2) - 20y = -64
(x^2) - 8x + (y^2) - 5y = 2.75

Reply 2

Christophicus

ryan750

4(x-4)^2 + (2y -5)^2 = 25

and

(x-4)^2 + (y - 2.5)^2 = 25

and if so - can any 1 explain pls?

4(x-4)^2 + (2y -5)^2 = 25
= 4x^2 - 32x + 64 + 4y^2 - 20y + 25 = 25
=> 4y^2 = 20y - 4x^2 + 32x - 64

(x-4)^2 + (y - 2.5)^2 = 25
x^2 - 8x + 16 + y^2 - 5y + 6.25 = 25
y^2 = 5y + 25 - 6.25 - x^2 + 8x - 16
4y^2 = 20y + 100 - 25 - 4x^2 + 32x - 64
4y^2 = 20y - 4x^2 + 32x + 11

Nope, they aren't.

Reply 3

Christophicus

By the way, el stevo, I think your avatar is really good.
Is it true that there is a professor of chemistry at oxford who has written over 400 chemistry books?

Reply 4

ryan750

i didnt think they were - but i wasnt sure - thanks for clarifying that. i will get some juicy questions for u guys soon probly when i get to a harder part.

Reply 5

Archkinkist

ryan750

4(x-4)^2 + (2y -5)^2 = 25

and

(x-4)^2 + (y - 2.5)^2 = 25

and if so - can any 1 explain pls?

Let's go through it
Part One,
4(x-4)(x-4)+(2y-5)(2y-5)=25.
==>4x^2-32x+64 +4y^2-20y+25=25.
Take 5 from each side, gives
4x^2-32x+64 +4y^2-20y+20=20
Common factor of 4 in every term so divide the whole thing by 4
x^2-8x+16+y^2-5y+5=5.

Second part;
(x-4)(x-4)+(y-2.5)(y-2.5)=25.
Multiplying out
x^2-8x+16 *the bit above* +y^2-5y+6.25=25. *similar to the bit above*
Since you'd be adding the same end values in both equations, to the number bits (...+6.25)=25 and (...+25)=25 (in this bit just add 20 on again afterwards to both sides to get equivalent second halves), then the two equations cannot be equal.
:eek: