SOLUTION: I need help solving the following for y in terms of x: x = y^4 + y^2 + 1 I can't think of much that I can do to it that is useful. What I have considered is that: x

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: I need help solving the following for y in terms of x: x = y^4 + y^2 + 1 I can't think of much that I can do to it that is useful. What I have considered is that: x      Log On


   

Question 143815: I need help solving the following for y in terms of x:

x = y^4 + y^2 + 1

I can't think of much that I can do to it that is useful. What I have considered is that:

x-1 = y^4 + y^2 = y^2(y^2 + 1)

and I could make a substitution u=y^2 so:

x-1 = u(u+1)

but again, I don't see the value in it if there is any. Perhaps it would help if I put this in context. This is from a calculus problem where I am to find dy/dx in terms of x. The idea is to find:

dx/dy = 1/[4y^3 + 2y] or [4y^3 + 2y]^(-1) and subsitute the y in terms of x which we solved above. As you can see, the difficult part is the algebra solving for y in terms of x. Any help would be much appreciated.

Greg
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
x+=+y%5E4+%2B+y%5E2+%2B+1 Start with the given equation

Let u=y%5E2. So u%5E2=y%5E4


x+=+u%5E2+%2B+u+%2B+1 Plug in u=y%5E2 and u%5E2=y%5E4



x+=+%28u%2B1%2F2%29%5E2%2B3%2F4 Complete the square to convert u%5E2+%2B+u+%2B+1 into %28u%2B1%2F2%29%5E2%2B3%2F4 (if you need help completing the square, check out this solver)


x-3%2F4+=+%28u%2B1%2F2%29%5E2 Subtract 3%2F4 from both sides


sqrt%28x-3%2F4%29+=+u%2B1%2F2 Take the square root of both sides


sqrt%28x-3%2F4%29-1%2F2+=+u Subtract 1%2F2 from both sides


u=sqrt%28x-3%2F4%29-1%2F2 Rearrange the terms


y%5E2=sqrt%28x-3%2F4%29-1%2F2 Replace "u" with y%5E2


y=sqrt%28sqrt%28x-3%2F4%29-1%2F2%29 Take the square root of both sides




So x+=+y%5E4+%2B+y%5E2+%2B+1 transforms into y=sqrt%28sqrt%28x-3%2F4%29-1%2F2%29