Quadratic, Cubic, Quartic Equations Solver
Solutions x1=


x2=
x3=
x4=
Input Equation:
First Derivation:
Second Derivation:
Extreme points x Value y Value position
1
2
3
Quadratic equation is of the form:        f(x) = ax2 + bx + c = 0
This equation has two solutions defined by:
The value under the root:    b2 - 4ac   is called the discriminant (Δ).
When   Δ > 0   then two real solutions exists.
When   Δ = 0   then one real solution exists.
When   Δ < 0   then two complex solutions exists.
The roots are related to each other by the formulas:
In order to find the extreme points of the graph we have to
find the first derivation of the function and compare it to 0.
To find if the extreme point is a maximum or minimum of
the graph we have to find the second derivation of the function.
If   f '' > 0   then the extreme point is a minimum.
If   f '' < 0   then the extreme point is a maximum.
Example 1: Find the extreme point of the function     f(x) = x2 + 3x - 10 = 0
The solutions of the function are:
First derivation is:    f ' = 2x + 3 = 0    extrime point at   x = -1.5
Second derivation is:   f '' = 2   
because   f '' >0   the extreme point is a minimum.
Example 2: Find the extreme points of the function     f(x) = 4x4 + 4x3 - 11x2 + 2x + 10 = 0
The solutions of the function are: -0.849,   -2.139,   0.994 ± i0.623
First derivation is:    f ' = 16x3 + 12x2 - 22x + 2 = 0
    (This curve is sketched by the red line)
The solutions of first derivation will give us the extreme points
which are at:    x1 = -1.64,    x2 = 0.1    x3 = 0.79
In order to find if the points are a maximum or minimum we
shall find the second derivation:       f ''(x) = 48x2 + 24x - 22
Now substitute each extreme point to the second derivation:
     f '' (x1) = 48 * (-1.64)2 + 24 * (-1.64) - 22 = 67.7
     f '' (x2) = 48 * 0.12 + 24 * 0.1 - 22 = -19.1
     f '' (x2) = 48 * 0.792 + 24 * 0.79 - 22 = 26.9
   x1 - is a minimum,    x2 - is a maximum,    and x3 - is a minimum