Please help me?

0 ⤊

Solve for y.
3x^2/y=2y+b . Thank you.

2007-08-26 04:21:06 · 4 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics

y is not 0 then
3x²=2y²+by
3/2x²+b²/16=y² +b/2y + b²/16
3/2x²+b²/16=(y+b/4)²
(-b/4)±√(3/2x²+b²/16)=y

2007-08-26 04:28:52 · answer #1 · answered by chasrmck 6 · 0⤊ 0⤋

This is actually a quadratic in y

3x^2 - 2y^2 - by = 0

Now solve for y assuming 3x^2 to be constant.

2007-08-26 11:29:53 · answer #2 · answered by ag_iitkgp 7 · 0⤊ 0⤋

3x^2=2 y^2+b y
if b=0 then
3x^2=2 y^2
x= y sqrt(2/3)
.
.

2007-08-26 11:32:54 · answer #3 · answered by Tuncay U 6 · 0⤊ 0⤋

It's hard to read the actual equation but it looks like you have three unknowns (x, y and b) and only one equation - in which case it can't be done.

2007-08-26 11:33:02 · answer #4 · answered by oracmike2001 1 · 0⤊ 0⤋