Finding all complex solutions to this problem

  • Thread starter Thread starter Cornraker
  • Start date Start date
  • Tags Tags
    Complex
AI Thread Summary
To solve the equation 4x^4 + 25x^2 + 36 = 0, substituting u = x^2 simplifies the problem to a quadratic equation in terms of u. The quadratic formula can then be applied to find the values of u. It is incorrect to factor out x^2 from the equation since it is not a common term across all parts. After solving for u, the final step involves taking the square root to find the complex solutions for x. This method effectively addresses the problem while ensuring all solutions are found.
Cornraker
Messages
24
Reaction score
0

Homework Statement




Find all solutions of the equation.

Homework Equations



4x^4+25x^2+36=0


The Attempt at a Solution



Where do i start here? do i use the quadratic formula? I'm thinking i can factor out a x^2 to make it{x^2(4x^2+25x+36) }and from there i can do the quadratic formula. Another question: can i not factor out x^2 since it is not common to all terms because 36 has no x^2?
 
Physics news on Phys.org
No, you can't factor x^2 out. You can, on the other hand, substitute u=x^2, solve for u using the quadratic equation and then find x.
 
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...
Essentially I just have this problem that I'm stuck on, on a sheet about complex numbers: Show that, for ##|r|<1,## $$1+r\cos(x)+r^2\cos(2x)+r^3\cos(3x)...=\frac{1-r\cos(x)}{1-2r\cos(x)+r^2}$$ My first thought was to express it as a geometric series, where the real part of the sum of the series would be the series you see above: $$1+re^{ix}+r^2e^{2ix}+r^3e^{3ix}...$$ The sum of this series is just: $$\frac{(re^{ix})^n-1}{re^{ix} - 1}$$ I'm having some trouble trying to figure out what to...
Back
Top