The sign of a quadratic trinomial

A ball is thrown upward. For which seconds is it more than $15$ m above the ground? "More than" is an inequality, and the height is a quadratic in time. To answer, we need to know where a quadratic is positive and where it is negative. The parabola tells the whole story.

The sign theorem

Picture it: for $a>0$ the parabola opens upward, so it lies above the $x$-axis wherever it does not dip below it. The sign changes only at roots.

Solving quadratic inequalities

Method: (1) move everything to one side; (2) find $\Delta$ and the roots; (3) use the theorem; (4) read off the answer.

A quotient inequality

Inequalities with a parameter

A word problem

Key terms

Summary

• Sign of $f(x)=ax^2+bx+c$: $\Delta<0$ gives the sign of $a$ everywhere; $\Delta=0$ gives the sign of $a$ except at the root; $\Delta>0$ gives the sign of $a$ outside the roots and the opposite sign between them.
• Solve an inequality by: one side zero, $\Delta$ and roots, then the theorem.
• For $a>0$: $f\ge 0$ for all $x$ $\iff\Delta\le 0$. For $a<0$: $f\le 0$ for all $x$ $\iff\Delta\le 0$.

Practice

Show solutions

1. Roots $1$ and $3$, $a>0$: positive outside. $S=(-\infty, 1)\cup(3, +\infty)$.
2. $x^2-9=(x-3)(x+3)$, roots $\pm3$, $a>0$: $\le 0$ between (including the roots). $S=[-3, 3]$.
3. $-x^2+2x+8=-(x-4)(x+2)$, roots $-2$ and $4$, $a<0$: positive between. $S=(-2, 4)$.
4. $2x^2-x-3=(2x-3)(x+1)$, roots $-1$ and $\frac32$, $a>0$: $\ge 0$ outside, with the roots. $S=(-\infty, -1]\cup\left[\frac32, +\infty\right)$.
5. $\Delta=16-20=-4<0$, $a>0$: always positive. $S=\mathbb{R}$.
6. $(x-2)^2\le 0$ only at $x=2$. $S=\{2\}$.
7. $a=1>0$, need $\Delta=m^2-16<0$, so $-4<m<4$.
8. Critical values $-1$ (numerator) and $3$ (excluded). Test: $x<-1$: $(-)/(-)>0$; $-1<x<3$: $(+)/(-)<0$ ✓; $x>3$: $(+)/(+)>0$. $S=(-1, 3)$.
9. Need $a=m<0$ and $\Delta'=4-m(m-3)<0$, i.e. $m^2-3m-4>0$, i.e. $(m-4)(m+1)>0$, so $m<-1$ or $m>4$. With $m<0$: $m<-1$. $m\in(-\infty, -1)$.
10. Need $x^2-3x-4\ge 0$: $(x-4)(x+1)\ge 0$, so $x\le -1$ or $x\ge 4$. Domain $(-\infty, -1]\cup[4, +\infty)$.