Why is $b^2-4ac<0$ if a linear line and a curve do not meet?

Question

For example there's a curve $y=X^2-4X+7$ and a line $L: Y=mX-2$. Both never intersect at any given point. If we were to find the set of values of $m$ for which $L$ does not meet the curve, the solution will be as follows

$$mX-2=X^2-4X+7$$

$$X^2-4X-mX+2+7=0$$

$$X^2-(4+m)X+9=0$$

Now here my teacher sets the inequality $b^2-4ac<0$.

$$(4+m)^2-36<0 $$

$$(4+m)^2<36$$

$$-6<(4+m)<6$$

$$-10<m<2$$

My question is why when a curve and a line do not intersect each other then the above inequality is set to further solve it. Is $b^2-4ac<0$ when two lines don't meet?

Please also explain whether this inequality is set only when a curve and a line do not meet.

Answer 1

It's because setting $b^2-4ac < 0$ means that the quadratic equation has no real solution. As for why this is the case, you most likely know the Quadratic Formula:

If $ax^2 + bx + c = 0$ and $a\neq 0$, then the two roots of the quadratic are found by $$x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}$$

Clearly, there will be no real solutions when the radicand or discriminant, $b^2-4ac$, is a negative value. Similarly, if one solution is required, the discriminant must equal $0$ (the curve and line intersect at one point only), and if two are required, then it must be a positive value (the curve and the line intersect at two points).

In the context of the given question, this means the line and the parabola never intersect at any point.