The normal form of a quadratic inequality is

(1)

(2)

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 $ax^2\!+\!bx\!+\!c = 0$ .

• 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 $$x < x_1\,\,\,\mbox{or}\,\,\,x > x_2$$ (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

• 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 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

• 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