Trying to analyze degenerate case of quadratic equation $ax^2+bx+c=0$ when $a=0,b=0,c \ne 0$

Question

Trying to analyze degenerate case of quadratic equation $ax^2+bx+c=0$ when $a=0,b=0,c\ne 0$

My attempt is as follows:-

$$x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$$

Let's take first root

$$x=\lim_{a\to 0,b\to 0}\dfrac{-b+\sqrt{b^2-4ac}}{2a}$$ $$x=\lim_{a\to 0,b\to 0}\dfrac{4ac}{2a\left(-b-\sqrt{b^2-4ac}\right)}$$

$$x=\lim_{a\to 0,b\to 0}\dfrac{2c}{\left(-b-\sqrt{b^2-4ac}\right)}$$

In the numerator we have finite value and denominator tends to zero and is negative, so limit is $-\infty$

Let's take second root

$$x=\lim_{a\to 0,b\to 0}\dfrac{-b-\sqrt{b^2-4ac}}{2a}$$ $$x=\lim_{a\to 0,b\to 0}\dfrac{4ac}{2a\left(-b+\sqrt{b^2-4ac}\right)}$$ $$x=\lim_{a\to 0,b\to 0}\dfrac{2c}{\left(-b+\sqrt{b^2-4ac}\right)}$$

In the numerator we have finite value and denominator tends to zero and is positive, so limit is $\infty$

So for example, this indicates that $f(x)=0x^2+0x+100$ meets the $x-$axis at $\infty$ and $-\infty$. But how it can be, how can it ever meet $x-$axis at infinity also?

Answer 1

The limits do not exist unless c=0 which is consistent with the degenerate quadratic equation.

Answer 2

Set $x=\dfrac1y$ to find $$0=a+by+cy^2$$

What if $a=0?$

What if $b=0?$

Now if $a=b=0,c\ne0,y^2=0$

Another point check where do the following two lines meet?

$$ax+by+c=0$$

$$ax+by+d=0$$

Answer 3

Consider two curves $y=ax+b$ and $y=-\dfrac{c}{x}$ in Cartesian plane.
Now as $a,b$ tends to zero, first curve (line) approaches to $x$-axis. On the other hand the (rectangular) hyperbola meets the $x$-axis at point at infinity. If you know little bit of protective geometry this would be much clear.