Need a little help with biquadratic equations

  • Thread starter Thread starter Noiro
  • Start date Start date
AI Thread Summary
The discussion focuses on proving that for the biquadratic equation x^4 + px^2 + q = 0, the sum of all solutions equals zero and the product equals q. The equation can be transformed into a quadratic form by substituting u = x^2, leading to u^2 + pu + q = 0. By applying the quadratic formula to find values of u, and then substituting back to find values of x, the relationships between the coefficients and roots can be explored. The conversation also suggests using polynomial expansion to relate coefficients to roots. Understanding these relationships is key to solving the problem effectively.
Noiro
Messages
24
Reaction score
0

Homework Statement



Ok, so here's the problem:
Prove, that biquadratic equation's x4+px2+q=0 sum of all solutions is equal to zero, and product is equal to q.

Homework Equations





The Attempt at a Solution


I don't really know how to solve it so I'd be very grateful if someone could help me. Thank you
 
Physics news on Phys.org
This equation is quadratic in form, which you can see by letting u = x2. This makes the new equation u2 +pu + q = 0.
Solve that equation for u by using the quadratic formula, getting two values of u.
Replace u by x2, getting four values of x.
 
While that equation is quadratic in form, for the purpose of the question think of it as x4+ 0*x3+ px2 +0*x+q=0. Do you know any general ways to relate the coefficients of a polynomial with the roots? If you're having trouble try multiplying out (x-a)(x-b)(x-c)(x-d)=0. That will become a polynomial with roots equal to a,b,c, and d.
 
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