Quadratic, Cubic, Quartic Equations calculator Print quadratic, cubic, quartic equations calculator
    
Solutions x1=




x2=
x3=
x4=
Input Equation:
First Derivation:
Second Derivation:
Extreme points x Value y Value position
1
2
3
First root x1 Quadratic equation is:
Second root x2
Quadratic, Cubic, Quartic Equations Notes Print quadratic, cubic, quartic equations notes
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 exist.
When   Δ = 0   then one real solution exists.
When   Δ < 0   then two complex solutions exist.
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    extreme 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.
Example 3: Find the quadratic equation if we know that the roots are 2 and 3.

It is easy to find the solutions of the x from the square equation by the following steps.

x_(1,2)=(-b±√(b^2-4ac))/2a (1)

Now we can add and multiply the solutions to get the solutions as a function of the coefficients of the square equation.

x_1+x_2=(-b)/a  x_1 x_2=c/a (2)








(3)

Substitute given solutions into equations  (2)  and  (3)

b =(x1 + x2)a =(2 + 3)a = −5a

c = x1x2a = 2 ‧ 3a = 6a

If we chose a to be equal to 1 than the quadratic equation will be:       x25x + 6 = 0