Best way to find possible roots?

M

Mystic

New member
Joined
Mar 8, 2022
Messages
4
Hi,

What would be the best way to find number of roots with different b values? Should discriminant be tried to use or factoring or how would you find possible number of roots for different values of b in this equation:
f\left(x\right)=4x^4+4bx^2+b^2
. I was factoring it out for
\left(2x^2+b\right)\left(2x^2+b\right)=0
and then because
2x^2
must be positive at all real values except zero then if
if\ b\ =\ 0{,}\ one\ root\ if\ x\ is\ also\ 0\

if\ b\ <0{,}\ zero\ roots\ because\ 2x^2=-b

if\ b>0\ two\ roots


I am just learning these so I don't have a very good whole picture of these issues. So I am thinking mainly the best way to solve this and how to mark it as clearly mathematically as possible without words. That is why I ask more info here. Thank you for help.
 
Factoring is usually the best way when it's possible. So you have two factors of [imath]2x^2 + b[/imath]. You actually have four roots here, all of them complex. ([imath]x = \pm i \dfrac{b}{\sqrt{2}}[/imath], repeated twice.)

-Dan
 
Mystic said:
Hi,

What would be the best way to find number of roots with different b values? Should discriminant be tried to use or factoring or how would you find possible number of roots for different values of b in this equation:
f\left(x\right)=4x^4+4bx^2+b^2
. I was factoring it out for
\left(2x^2+b\right)\left(2x^2+b\right)=0
and then because
2x^2
must be positive at all real values except zero then if
if\ b\ =\ 0{,}\ one\ root\ if\ x\ is\ also\ 0\

if\ b\ <0{,}\ zero\ roots\ because\ 2x^2=-b

if\ b>0\ two\ roots


I am just learning these so I don't have a very good whole picture of these issues. So I am thinking mainly the best way to solve this and how to mark it as clearly mathematically as possible without words. That is why I ask more info here. Thank you for help.
Click to expand...
You're almost, right if you're looking for only real roots. If [imath]b=0[/imath], the equation is [imath]4x^4 = 0[/imath], so you have one root, namely 0; otherwise, the roots are those of [imath]2x^2 = -b[/imath], so there are two real roots, [imath]\pm\sqrt{-\frac{b}{2}}[/imath], if [imath]b<0[/imath], and two imaginary roots if [imath]b>0[/imath]. You got that condition backwards.
 
A polynomial of degree n will have n roots (including multiplicative).
Do you mean real roots? If so, then please state that.
Discriminants only really come up for quadratic equations. Are you only asking about quadratic equations??
 
If you are not interested in exact answers, graph the thing.

Even if you are interested in exact answers, graphing may give you clues about what might be rational roots or what might be exact irrational roots.
 
Thank you for your answers! Yes I was looking only ways to solve real roots.
 
You must log in or register to reply here.
Top