Transforming x^5-x^4-x^3-x^2-x-1 to y^5+2y^2+47y+122

  • Thread starter Thread starter footmath
  • Start date Start date
AI Thread Summary
To transform the equation x^5 - x^4 - x^3 - x^2 - x - 1 = 0 into y^5 + 2y^2 + 47y + 122, the substitution y = x^2 - 3x is suggested. However, the challenge arises when attempting to express x in terms of y, leading to the equation x = 3/2 + √(y + 9/4). Substituting this expression back into the original polynomial does not yield the desired form of y^5 + 2y^2 + 47y + 122. Clarification is sought on the correct method to achieve this transformation. Understanding the proper substitution process is crucial for solving the equation accurately.
footmath
Messages
26
Reaction score
0
Hello , please guide me .
How can I transformed the equation x^{5}-x^{4}-x^{3}-x^{2}-x-1 =0to y^{5}+2y^{2}+47y+122 ?
I studied a lecture that the writer had written :<< by using y=x^{2}-3x we can transformed x^{5}-x^{4}-x^{3}-x^{2}-x-1 =0 to y^{5}+2y^{2}+47y+122 But how ? if in y=x^2-3x we obtain x by y we will have : x=3/2+\sqrt{y+9/4} and if we substitute x=3/2+\sqrt{y+9/4} in x^{5}-x^{4}-x^{3}-x^{2}-x-1 =0 we don't have $y^5+2y^2+47y+122$ . please explain it.
Thank you very much
 
Physics news on Phys.org
footmath, please edit your post to fix your LaTeX. All of your TEX and /TEX tags need to be written in lower case - as tex and /tex (inside square brackets).
 
I picked up this problem from the Schaum's series book titled "College Mathematics" by Ayres/Schmidt. It is a solved problem in the book. But what surprised me was that the solution to this problem was given in one line without any explanation. I could, therefore, not understand how the given one-line solution was reached. The one-line solution in the book says: The equation is ##x \cos{\omega} +y \sin{\omega} - 5 = 0##, ##\omega## being the parameter. From my side, the only thing I could...
Back
Top