If $ax^2+bx+c$ is irreductible, then exists constants $k_1,k_2,k_3$ such that $ax^2+bx+c = k_1((k_2x+k_3)^2+1).$ I want to prove that:

If $ax^2+bx+c$ is irreductible, then exists constants $k_1,k_2,k_3$ such that
$$ax^2+bx+c = k_1((k_2x+k_3)^2+1)$$

We note that
\begin{align*}
&ax^2+bx+c = \frac{1}{4a}(4a^2x^2+4abx+4ac)= \frac{1}{4a}[(4a^2x^2+4abx+b^2)-b^2+4ac]=\\
&= \frac{1}{4a}[(2ax+b)^2-(b^2-4ac)]\\
\therefore\ ax^2+bx+c &= \frac{1}{4a}[(2ax+b)^2-(b^2-4ac)].
\end{align*}
Now, $ax^2+bx+c$ is irreductible implies that $b^2-4ac<0$, since otherwise we have that $ax^2+bx+c$ has roots and in consequence it can be expressed as something of the form $(x-r_1)(x-r_2)$, where $r_1,r_2\in\mathbb{R}$, and in consequence it is not irreductible.
I think thta is necesary that $b^2-4ac=-1$ but I can't see why, can anyone help me please?

$$
q(x) = \frac{1}{4a} ((2ax+b)^2 - (b^2 - 4ac) )\\
D = b^2 - 4ac < 0\\
r \equiv \sqrt{-D}\\
q(x) = \frac{1}{4a} ((2ax+b)^2 + r^2 )\\
= \frac{r^2}{4a} (\frac{(2ax+b)^2}{r^2} + 1 )\\
= \frac{r^2}{4a} ( (\frac{2a}{r} x + \frac{b}{r})^2 + 1)\\
k_1 = \frac{r^2}{4a}\\
k_2 = \frac{2a}{r}\\
k_3 = \frac{b}{r}\\
$$
Note that you had to take a square root of a positive number. This is allowed if you just wanted to have $k_i$ be real numbers. But if you said for example, $a,b,c$ were rational and you wanted $k_i$ to also be rational this would not work. But since this is about precalculus, I am guessing you only care about numbers as real numbers not whether they are more specifically rational, integers, etc.
