Quartic Equation Solution and Conditions for real roots? Q1. How to solve a Quartic Equation. There is an online calculator available (and many more similar) that gives the precise answers and also defines the method. Does anyone know what the source of this method is?
Q2. Given a Quartic Equation 
$$
ax^4+bx^3+cx^2+dx+e=0\,,
$$
what are the conditions for the existence of real roots of the above equation?
Any reference material?
 A: What about a set of expressions in the quartic's coefficients that discriminate between all cases?
There are 9 cases:


*

*4 distinct real roots.

*3 distinct real roots with one of them being a double root.

*2 distinct double roots both real.

*Triple root and and a distinct fourth root.

*Quadruple root.

*2 distinct real roots and two complex roots.

*double real root and 2 complex roots.

*2 double roots both complex.

*four distinct complex roots.
Such sets are known for quadratic and cubic polynomials:
Quadratic ($ ax^2+bx+c $):
Discriminant: $b^2-4ac$
Positive for two distinct real roots, zero for double root, and negative for complex conjugate roots.
Cubic ($ ax^3+bx^2+cx+d $):
$\Delta_1=2b^3-9abc+27a^2d$ and $\Delta_2=\Delta_1^2-4(b^2-3ac)^3$.
Then:
$ \Delta_2>0 $ gives one real root and two complex roots.
$ \Delta_2<0$ gives three distinct real roots.
$ \Delta_2=0 $ but $ \Delta_1\neq 0$ gives a double root plus one different root.
$ \Delta_1=\Delta_2=0$ gives a triple root.
A: There is no simple search method that always works for the count or estimatica roots. In general it is a rough approximation $a$ of the root $x$ is used the bisection method. And it applies the method of$x_0=a\quad x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}$ Newton for the bestapproximation of root. That is, $\lim_{n\to \infty}x_n=root$. On more times, $|x_{10}-root|<0.01$.
Another method of counting real roots escartes beyond the rule and that is easy to see geometrically is to rank the extreme points of the equation with maximum, minimum and saddle point.
1) We know the fundamental theorem algebra that the number of roots of this equation is at most equal to four.
2)Let $ x $ is a point that extreme point ($ f '(x) = 0 $):


If $ x $ is the point of minimum ($ f'' (x)> 0 $) and satisfies $ f (x)> 0 $ then the equation has two complex roots.


 If $ x $ is the maximum point ($ f'' (x) <0 $) and satisfies $ f (x) <0 $ then the equation has two complex roots.

If $ x $ is the point of cell $ f'' (x) = 0 $ and $ f (x) = 0 $ then $ x $ is double root of the equation.

3) You can also use the Intermediate Value Theorem ([See wikipedia][1]) to verify the existence of roots in a reiais interestimar the interval $ [a, b] $. In fact, if $f(a)<0<f(b)$ ou $f(b)>0>f(a)$, then there is a x ∈ [a, b] such that f(x) = 0.
A: MATLAB function to solve the general case quartic equation (based on the full formula from Wikipedia). Unfortunately, it does not determine which roots are real (I am working on that now).
function [r1, r2, r3, r4]=quarticsolve(a4,a3,a2,a1,a0)
%solving quartic equation

    a=a3/a4;
    b=a2/a4;
    c=a1/a4;
    d=a0/a4;

    third=1.0/3.0;
    root3=2^third;

    q1=(b*b-3*a*c+12*d);
    q2=2*b*b*b-9*a*b*c+27*c*c+27*a*a*d-72*b*d;
    q3=-4*q1*q1*q1+q2*q2;
    q4=-a*a*a+4*a*b-8*c;

    s0=0.25*a*a-2*b*third;
    s1=root3*q1*third/(q2+sqrt(q3))^third;
    s2=((q2+sqrt(q3))/54)^third;

    r1=-0.25*a-0.5*sqrt(s0+s1+s2)-0.5*sqrt(2*s0-s1-s2-0.25*q4/sqrt(s0+s1+s2));
    r2=-0.25*a-0.5*sqrt(s0+s1+s2)+0.5*sqrt(2*s0-s1-s2-0.25*q4/sqrt(s0+s1+s2));
    r3=-0.25*a+0.5*sqrt(s0+s1+s2)-0.5*sqrt(2*s0-s1-s2+0.25*q4/sqrt(s0+s1+s2));
    r4=-0.25*a+0.5*sqrt(s0+s1+s2)+0.5*sqrt(2*s0-s1-s2+0.25*q4/sqrt(s0+s1+s2));
end

A: It suffices to solve the equation
$$x^{4}+ax^{2}+bx+c=0.$$
It is possible to factor this polynomial into two polynomials of degree 2.
In order for us to have zero coefficient for $x^{3}$ , and the free term to be equal to c, we must have a factorization of the form
$$x^{4}+ax^{2}+bx+c=\left(x^{2}+\alpha x+\beta\right)\left(x^{2}-\alpha x +\frac{c}{\beta}\right)=0.$$
Equating the coefficients we get
$$a=\frac{c}{\beta}+\beta-\alpha^{2}$$
$$b/\alpha=\frac{c}{\beta}-\beta.$$
or
$$\alpha^{2}+a=\frac{c}{\beta}+\beta$$
$$\frac{b}{\alpha}=\frac{c}{\beta}-\beta.$$
Note that we can eliminate $\beta$ by using the following trick
$$(\alpha^{2}+a)^2-\left(\frac{b}{\alpha}\right)^2=4c$$
substituting $t=\alpha^2$ we obtain the following cubic equation for $t$:
$$t(t+a)^2-4ct-b^{2}=0.$$
Then solving the cubic equation, and selecting any solution we can deduce the values of $\alpha$. Then we need to solve the quadratic equation
$$\beta^{2}+\frac{\beta}{\alpha}\beta-c=0.$$
Solving the equation and selecting any of the possible values for $\alpha$ and $\beta$, solving the quartic reduces to solving two quadratic equations
$$x^{2}+\alpha x+\beta=0$$ and  $$x^{2}-\alpha x +\frac{c}{\beta}=0.$$
A link to a detailed video lecture is also available.
https://www.youtube.com/watch?v=g0J3v0mQItc&ab_channel=Math%2CPhysics%2CEngineering
