Math question

[Scrapster]

Here's the equation:

3y + y^2 = x^3 -x + c

c is constant.

Goal: solve for y.

What is the simplest way to solve for y?

Any ideas?

Scrapster

 

[Unsickle]

quadratic equation Treat the right hand side as a constant C and then resubstitute once solved.

 

[Nutcase99]

Duhhhh? Me no know math

 

[luckydragon]

me no english

 

[Scrapster]

Unsickle,

Can you ellaborate more. Maybe work through the first few steps?

Scrapster

 

[blurredvision]

Why don't you just ASK him to do your homework for you??

I came up with y=mc squared.......

 

[Scrapster]

Woody,

Because I'm already done with half the problem. I'm stuck on this one part. It's the weekend so no tutors or instructors are around.

Scrapster

 

[Mister T]

y1 = (-3 + squareroot(4x^3-4x+4c+9))/2
y2 = (-3 - squareroot(4x^3-4x+4c+9))/2

 

[Scrapster]

Ok. I see the answer. But I don't understand how you got it.

Can someone please give me steps on how I can solve this problem?

Scrapster

 

[Bluga]

Let (x^3 -x + c) = K ( K is constant)

so 3y + y^2 =K

==> y^2 +3y -K = 0

use equation y = [-b (+-) squareroot(b^2-4ac)] /2a

then substitue

 

[Scrapster]

I got it.

You guys were right and I just wasn't following well. I didn't think you could substitute (x^3 -x + c) as your quadratic constant. But once you do that, it's simple enough. I must have been sleeping that day in algebra class.



Scrapster