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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?

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Plese see. mathworld.wolfram.com/DescartesSignRule.html. –  Elias Mar 2 '12 at 11:30
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You still haven't read this paper? I notice that you asked this months ago... –  J. M. Apr 14 '12 at 5:13
    
See also: math.stackexchange.com/a/786/72 –  Isaac May 2 '12 at 7:16
    
math.stackexchange.com/users/21481/abhinav math.stackexchange.com/users/26124/abhinav Are both of these your account? –  user17762 May 14 '12 at 18:23
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2 Answers

What about a set of expressions in the quartic's coefficients that discriminate between all cases? There are 9 cases:

  1. 4 distinct real roots.

  2. 3 distinct real roots with one of them being a double root.

  3. 2 distinct double roots both real.

  4. Triple root and and a distinct fourth root.

  5. Quadruple root.

  6. 2 distinct real roots and two complex roots.

  7. double real root and 2 complex roots.

  8. 2 double roots both complex.

  9. 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.

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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.

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the method that online calculator describes is not iterative and is strangely analytic. any ideas how the calculator is obtaining roots of a quartic equation directly just like solving a quadratic equation without computing any iterations? –  Abhinav Mar 3 '12 at 5:32
    
The method which the calculator applies is reported at 1728.org/quartic2.htm –  Abhinav Mar 3 '12 at 5:34
    
Then see en.wikipedia.org/wiki/Quartic_function –  Elias Nov 12 '12 at 19:50
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