quadratic inequality

$\displaystyle ax^{2}\!+\!bx\!+\!c\;<\;0$

(1)

$\displaystyle ax^{2}\!+\!bx\!+\!c\;>\;0$

(2)

where $a$ , $b$ and $c$ are known real numbers and $a\neq 0$ .

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 $x\mapsto ax^{2}\!+\!bx\!+\!c$ is the parabola

$y\;=\;ax^{2}\!+\!bx\!+\!c,$

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 $ax^{2}\!+\!bx\!+\!c$ , i.e. solve the quadratic equation (http://planetmath.org/QuadraticFormula) $ax^{2}\!+\!bx\!+\!c=0$ .

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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

$x\;<\;x_{1}\,\,\,\mbox{ or }\,\,\,x\;>\;x_{2}$

(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 $ax^{2}\!+\!bx\!+\!c<0$ when $a>0$ and the quadratic has two distinct roots

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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 $x\neq x_{1}$ (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 $ax^{2}\!+\!bx\!+\!c>0$ when $a>0$ and the quadratic has only one root
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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: $ax^{2}\!+\!bx\!+\!c>0$ 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