icon-Mathe Quadratic equation

Calculator

Computer for the solution of the quadratic equation:

a·x2+b·x+c = 0

Coefficients

Inputting the coefficients a, b and c

a= b= c=

Graphical Representation

Graphical representation of the associated quadratic function (parabola) f(x).

fx = a·x2+b·x+c

x-min= x-max=

Quadratic Equation

Basics

Basic form of the quadratic equation with constant coefficients a, b and c:

a·x2+b·x+c = 0

witha,b,cRunda0

Normal form

Division by the coefficients a and renaming of terms ba and ca leads to the normal form of the quadratic equation:

x2+bax+ca = 0

withp=baundq=cafollows the basic form

x2+px+q = 0

General solution of the quadratic equation

By means of forming and application completing the square, the general solution of the quadratic equation in the form of p, q-formula are given:

x2+px+q = 0

x2+px = -q

x2+px+p22-p22 = -q

x2+px+p22 = p22-q

x+p22 = p22-q

x+p2 = ±p22-q

x1,2 = -p2±p22-q

Starting from the normal form of the quadratic equation, the equation is solved using quadratic supplement.

Starting point for the general solution is the normal form of the quadratic equation.

1. Subtraction q

2. Enhancements to the equation p22 and subtraction of this term, so that the equation is not actually changed.

3. After the conversion is on the left side of the equation a term that corresponds to the first binomial theorem: a+b2=a2+2ab+b2

4. Application of the binomial leads to a quadratic expression.

5. Pull the root then allows the resolution of the equation for x. Because the square root in general has a positive and a negative quadratic equation, the solution also in general two solutions x1 und x2.

6. Result is the so-called. p,q-formula for determining the solution of a quadratic equation.

The solutions can be divided into three categories depending on the value of the discriminant: D=p22-q :

D=0 : There is one real solution.

D>0 : There are two real solutions.

D<0 : There are two complex solutions.

Examples

Example of a quadratic equation with two real solutions

The first example has two real solutions. In the following, the approach is shown with a square expansion and then with the pq-formula.

x2+3x+2 = 0

Example equation

x2+3x = -2

Subtracting the absolute term

x2+3x+322-322 = -2

With the addition of the term 322 the expression is extended to the first binomial formula.

x2+232x+322-322 = -2

Expanding the factor in front of x illustrates the binomial structure.

x2+232x+322 =322 -2

After forming is on the left side of the equation the first binomial.

x+322 =322 -2

Application of the first binomial theorem a+b2=a2+2ab+b2

x+32 =±322 -2

Application of the square root allows the resolution of the equation for x. The square root of generally has a positive and a negative solution.

x1,2 =-32±322 -2

x1 =-1

x2 =-2

Forming and calculating results on the two real solutions of the quadratic equation.

x2+3x+2 = 0

To the solution reaches you even by employing the coefficients of the equation in the p,q-formula.

x1,2 = -p2±p22-q

p and q must be replaced by the coefficients.

x1,2 =-32±322 -2

x1 =-1

x2 =-2

Use of p = 3 and q = 2 gives the solution of the equation.

Example of a quadratic equation with two complex solutions

The second example has two complex solutions. In the following, the approach is first with square complement and then shown with the p/q-formula.

x2+1 = 0

Example equation

x2 = -1

This very simple quadratic equation can be transformed directly.

x1,2 = -1 = ±i

x1 =+i

x2 =-i

The special feature is that the discriminant is negative so the term under the square root. The square root of -1 is denoted by i. The i stands for imaginary unit.

x2+0x+1 = 0

To the solution is arrived at by substituting the coefficients of the equation in the p,q-formula.

x1,2 = -p2±p22-q

p and q must be replaced by the coefficients.

x1,2 =-02±022 -1

x1 =+i

x2 =-i

Inserting p = 0 and q = 1 results in the solution of the equation.

Example of a quadratic equation with a two-fold solution

The third example has a two-fold real solution.

x2+4x+4 = 0

Example equation

x2+4x+4-4 = -4

x+22 = 0

x1,2 =-2

The solution by completing the square leads to a discriminant with a value of 0 means that there is a twofold solution before with +/- 0th

x+2x+2 = x2+4x+4

x1,2 =-2

From the product representation of the equation it is apparent that it is a two-fold solution.

The p,q-formula for solving a quadratic equation

The application of p,q formula requires that the quadratic equation is in the normal form. If they do not exist so they can be converted by transformations in the normal form. Here an example the necessary transformations to normal form.

2x2-4x+6 = 2

Example equation

x2-2x+3 = 1

Division by the factor before x2

x2-2x+2 = 0

Subtracting the right-hand side

x2+-2x+2 = 0

Taking into account the sign of p = 2 can be read off p = -2 and q.

Parabola

The solutions of the quadratic equation corresponding to the zeros of a parabola. A parabola is defined by a mapping of the form correspond to the zeros of the function fx=a·x2+b·x+c . From this follows that the solution of the quadratic equation a·x2+b·x+c=0 corresponds to the zeros of the function f(x). Where the parabola intersects the x-axis are the solutions to the equation.

Parabel

Depending on the location of the parabola are two zeros, one zero or no zeros. If the parabola do not intersect the x-axis has the corresponding quadratic equation complex solutions.

Parabola Plotter

Interactive graphical representation of a parabolaParabola PlotterFunction_plot