opening upwards if and downwards if .
If there is two distinct real zeroes and (say ), then the parabola intersects the -axis in these points. In the case the parabola opens upwards and thus in the interval , but outside this interval. I.e., for positive , the solution of (1) is
and the solution of (2) is
(note that the latter solution-domain consists of two distinct portions of the -axis and therefore must be expressed with two separate inequalities, not with a double inequality as the former). For negative we must swap those solutions for (1) and (2).
If there is only one real zero of the polynomial (we may say that ), the parabola has -axis as the tangent (http://planetmath.org/TangentLine) in its apex. For positive the other points of parabola are above the -axis, i.e. they have always but never. So, (1) has no solutions, but (2) is true for all (i.e. or ). For the case of negative we again must change those solutions for (1) and (2).
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 -axis, but is totally above the axis when is positive ( always) and totally below the axis when is negative ( always). Thus we get no solutions at all (the inequality is impossible) or all real numbers as solutions, according to what the inequality (1) or (2) demands.
|Date of creation||2013-03-22 15:23:48|
|Last modified on||2013-03-22 15:23:48|
|Last modified by||pahio (2872)|