Quadratic inequality — sign of the trinomial

What determines the sign of the trinomial?

For the quadratic function $f(x)=ax^2+bx+c$ with coefficient $a>0$, the graph is an upward-opening parabola. Whether it crosses the x-axis is decided by the discriminant:

The discriminant tells you the number of roots: $\Delta>0$ — two roots, $\Delta=0$ — one (double), $\Delta<0$ — no real roots.

Sign when a > 0:

When there are roots $x_1<x_2$, the trinomial is negative between them and positive outside them:

The key case — no roots:

If $\Delta<0$, the parabola never touches the x-axis, and with $a>0$ it lies entirely above it. Then:

Why this matters for the exam:

You solve quadratic inequalities by a routine: compute $\Delta$, find the roots, sketch the parabola and read off the sign. When $\Delta<0$ the answer is the empty set or the whole line — depending on the sign of $a$ and the direction of the inequality.

Remember: the sign of the trinomial depends on the sign of $a$ and the value of $\Delta$. Rule: $a>0 \wedge \Delta<0 \Rightarrow ax^2+bx+c>0$ for every x.