/sci/ - Science & Math

[Anonymous No.6007436]

I'm begging you guys to help me figure out what the fuck I'm missing on this retarded completing the square problem

Here's the problem

4x^2-3x+4y^2=8y-8=0
Here's what I'm doing
Add 8 to both sides and group like terms (already done)
4x^2-3x+4y^2+8y=8
Divide 4 from both sides so you end up with
x^2-3/4x+y^2+2y=2
Then you complete the square by taking the middle terms, dividing them by 2, squaring and adding to both sides, this is where I fall apart. I get
x^2+3/4x+(3/8)^2+y^2+2y+2^2=2+(3/8)^2+1^2

I need h, k and r. h and k are .375 and -1 respectively but for some reason, I can't figure out r. I think this problem may be wrong in my homework but I'm at a loss for places to look for help, the TA won't answer my email and I'm getting pretty desperate.

[Anonymous No.6007472]

bump

[Anonymous No.6007498]

Holy fuck I hate online homework the problem wasn't my answer, it was the way the stupid online form wanted it entered.

My rage is beyond apocalyptic, you could weld steel to butter with it right now.

[Anonymous No.6007540]

try again

[Anonymous No.6008793]

>>6007540
4x^2-3x+4y^2-8y+8=0 (adding -8y+8)
x^2-3/4x+y^2-2y+2=0 (dividing by 4)
(x-3/8)^2-(3/8)^2 + (y-1)^2-1+2=0 ( using a^2+2ab = (a+b)^2-b^2 )
(x-3/8)^2 + (y-1)^2=(3/8)^2-1
since 3/8<1, last term is <0
->first term is a sum of squared, so >=0

-> No solution

(if I didn't fuck up my calculus. If I did, the idea is to get in the end (x-a)^2+(y-b)^2=R^2
-> the solution is all the points on the cricle of center C(a,b) and radius R)