The of a quadratic inequality is

 $\displaystyle ax^{2}\!+\!bx\!+\!c\;<\;0$ (1)

or

 $\displaystyle ax^{2}\!+\!bx\!+\!c\;>\;0$ (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)  $ax^{2}\!+\!bx\!+\!c=0$.

• If there is two distinct real zeroes $x_{1}$ and $x_{2}$ (say  $x_{1}),  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 a⁢x2+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  $x\neq x_{1}$ (i.e.  $x  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⁢x2+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⁢x2+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