quadratic inequality

quadratic inequality

The of a

quadratic inequality

is

a

⁢

x

2

+

b

⁢

x

+

c

<

0

(1)

or

a

⁢

x

2

+

b

⁢

x

+

c

>

0

(2)

where

a

,

b

and

c

are known

real numbers

and

a

≠

0

.

Solving such an

inequality

, i.e. determining all real values of

x

which satisfy it, is based on the fact that the graph of the

quadratic polynomial

function

x

↦

a

⁢

x

2

+

b

⁢

x

+

c

is the parabola

y

=

a

⁢

x

2

+

b

⁢

x

+

c

,

opening upwards if

a

>

0

and downwards if

a

<

0

.

For obtaining the solution we first have to determine the real zeroes of the

polynomial

a

⁢

x

2

+

b

⁢

x

+

c

, i.e. solve the

quadratic equation

(

http://planetmath.org/QuadraticFormula

)

a

⁢

x

2

+

b

⁢

x

+

c

=

0

.

•

If there is two distinct real zeroes

x

1

and

x

2

(say

x

1

<

x

2

),  then the parabola intersects the

x

-axis in these points.  In the case

a

>

0

the parabola opens upwards and thus

y

<

0

in the

interval

(

x

1

,

x

2

)

, but

y

>

0

outside this interval.  I.e., for positive

a

, the solution of (1) is

x

1

<

x

<

x

2

and the solution of (2) is

x

⁢

<

x

1

⁢

or

⁢

x

>

⁢

x

2

(note that the latter solution-domain consists of two distinct portions of the

x

-axis and therefore must be expressed with two separate inequalities, not with a double inequality as the former).  For negative

a

we must swap those solutions for (1) and (2).

Figure 1:

Solving for

a

⁢

x

2

+

b

⁢

x

+

c

<

0

when

a

>

0

and the quadratic has two distinct roots

•

If there is only one real zero of the polynomial (we may say that

x

2

=

x

1

), the parabola has

x

-axis as the

tangent

(

http://planetmath.org/TangentLine

) in its apex.  For positive

a

the other points of parabola are above the

x

-axis, i.e. they have

y

>

0

always  but

y

<

0

never.  So, (1) has no solutions, but (2) is true for all

x

≠

x

1

(i.e.

x

<

x

1

or

x

>

x

1

).  For the case of negative

a

we again must change those solutions for (1) and (2).

Figure 2:

Solving for

a

⁢

x

2

+

b

⁢

x

+

c

>

0

when

a

>

0

and the quadratic has only one root

•

There can still appear the possibility that the polynomial has no real zeroes (the roots of the equation are

imaginary

).  Now the parabola does not intersect or touch the

x

-axis, but is totally above the axis when

a

is positive (

y

>

0

always) and totally below the axis when

a

is negative
(

y

<

0

always).  Thus we get no solutions at all (the inequality is impossible) or all real numbers

x

as solutions, according to what the inequality (1) or (2) demands.

Figure 3:

a

⁢

x

2

+

b

⁢

x

+

c

>

0

for all

x

when

a

>

0

and the quadratic has no roots

Title

quadratic inequality

Canonical name

QuadraticInequality

Date of creation

2013-03-22 15:23:48

Last modified on

2013-03-22 15:23:48

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

12

Author

pahio (2872)

Entry type

Topic

Classification

msc 97D40

Classification

msc 26-00

Classification

msc 12D99

Related topic

QuadraticFormula

Related topic

SolvingCertainPolynomialInequalities

Related topic

TangentOfConicSection

Related topic

IndexOfInequalities

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