Implicit Function Thm Application

[trap101]

Investigate the possibility of solving x²-4x+2y²-yz = 1 for each of its variables in terms of the other two near the point (2,-1,3).

Attempt:

Ok so using the IFT I was able to determine that I can only slze for y,z. But in the question they ask me to solve for y and z. z was not problem, but y, they got a solution like this:

y = (z - [z²+8(1-x²+4x)]^1/2) / 4

as well they said x = 2 (+ or -) [5-2y²+yz]^1/2 and the square root vanishes at (y,z) = (-1,3) so there are two values of “ for some nearby values of and
and none for others.

Question: How did they solve for y? (What bloody algebra trick did they use that I'm not seeing? and how did those conditions on x appear?

 

[tiny-tim]

hi trap101!

x²-4x+2y²-yz = 1
…
y = (z - [z²+8(1-x²+4x)]^1/2) / 4

that's the standard -b ± √etc formula

as well they said x = 2 (+ or -) [5-2y²+yz]^1/2 and the square root vanishes at (y,z) = (-1,3) so there are two values of “ for some nearby values of and
and none for others.

your typing hasn't come out

 

[trap101]

as well they said x = 2 (+ or -) [5-2y²+yz]^1/2 and the square root vanishes at (y,z) = (-1,3) so there are two values of “ for some nearby values of and ��
and none for others.

there are two values of x for some nearby values of y and z
and none for others.
Well after doing the mechanics for solving for y, I guess in that sort of scenario I should assume the other "variables" in this case x, z are just constants?

 

[tiny-tim]

Well after doing the mechanics for solving for y, I guess in that sort of scenario I should assume the other "variables" in this case x, z are just constants?

yup!

 

[trap101]

thnks again and won't be the last time, unless I get struck by lightning and all of a sudden I become a genius