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

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

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.

• That comes from the theory of quadratic equations, for we know exactly when the equation has no real solutions. – Andrea Mori Mar 2 at 17:38
• "If we were to find the set of values of $m$ for which $L$ does not meet the curve, [equate the Y in the two equations]" - note that that's what you to find where $L$ does meet the curve: at the point if intersection, both equations are satisfied with the same $(X, Y)$: the coordinates of the point of intersection. What you need to do to prove that there is no such point is to prove that the resulting equation has no (real) solutions. – NickD Mar 3 at 1:42

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).