Sign of a Quadratic Expression and How It Changes

For a fixed real quadratic $f(x)=ax^2+bx+c$ ($a,b,c$ real, $a\neq 0$), a natural question is: for which real $x$ is $f(x)$ positive, and for which is it negative? The discriminant answers this completely.

When $\Delta < 0$ (no real roots): $f(x)$ and the leading coefficient $a$ have the same sign for every real $x$, with no exceptions. This follows from completing the square: $f(x) = a\Big[\big(x+\frac{b}{2a}\big)^2 + \frac{4ac-b^2}{4a^2}\Big]$, and when $\Delta<0$ the bracket is always strictly positive, so $f(x)$ is a positive multiple of $a$ throughout. For instance $x^2+x+1$ has $\Delta = 1-4 = -3<0$ and $a=1>0$, so $x^2+x+1>0$ for every real $x$ — it never touches zero.

When $\Delta = 0$ (equal roots $-b/2a$): the same argument shows $f(x)$ and $a$ have the same sign everywhere except at $x=-\dfrac{b}{2a}$, where $f(x)=0$ exactly.

When $\Delta > 0$ (two distinct real roots $\alpha<\beta$): now $f(x)$ factorises as $a(x-\alpha)(x-\beta)$, and the sign flips depend on where $x$ sits relative to the roots:

• For $\alpha < x < \beta$ (strictly between the roots): $x-\alpha>0$ and $x-\beta<0$, so $(x-\alpha)(x-\beta)<0$ — the expression $f(x)$ and $a$ have opposite signs here.
• For $x<\alpha$ or $x>\beta$ (outside the roots): both factors have the same sign as each other, so $f(x)$ and $a$ have the same sign.

So the sign of a quadratic changes exactly at its real roots, and it changes back once you cross the second root — this "same–opposite–same" pattern (relative to the sign of $a$) is the single fact that makes solving quadratic inequations possible without ever drawing a graph. …