Firstly, observe that if $f(x), g(x)$ satisfy the condition that $f, g, f+g$ all have a discriminant of $1$, then letting $h(x) = -f(x) - g(x)$ satisfies all six listed conditions. We will show that no other polynomial $h(x)$ works. The six equations are $$b_1^2 = 4a_1c_1 + 1,$$$$b_2^2 = 4a_2c_2 + 1,$$$$b_3^2 = 4a_3c_3+1,$$$$4a_1c_2 + 4a_2c_1 = 2b_1b_2 + 1,$$$$4a_2c_3 + 4c_2a_3 = 2b_2b_3 + 1,$$$$4a_1c_3 + 4c_1a_3 = 2b_1b_3 + 1.$$ Let $f(x) = a_1x^2 + b_1x + c_1, g(x) = a_2x^2 + b_2x + c_2, h(x) = a_3x^2 + b_3x + c_3.$ Since they are trinomials, we know that none of the $a_i$'s is $0$. Then, the third, fifth, and sixth equations above give us a system of three equations in $a_3, b_3, c_3$ (note that $a_1, a_2, b_1, b_2, c_1, c_2$ are constants here). We'll show the uniqueness of $a_3, b_3, c_3$ in three cases. Case 1. $a_1 + a_2 + a_3 = 0.$ WLOG that $b_1, b_2, b_3, c_1, c_2, c_3 \neq 0$, as otherwise we can look at $f(x+c), g(x+c), h(x+c)$ instead for some suitable $c \in \mathbb{R}.$ (discriminant is preserved under horizontal shifts). Our three relevant conditions now read: $$4a_2c_3+ 4c_2a_3 = 2b_2b_3 + 1,$$$$4a_1c_3 + 4c_1a_3 = 2b_1b_3 + 1,$$$$b_3^2 = 4a_3c_3 + 1.$$Notice that $a_3$ is now a constant ($-a_1-a_2$). Summing the three equations gives us that: $$4c_2a_3 + 4c_1a_3 + 1 = b_3^2 + 2b_1b_3 + 2b_2b_3 + 2,$$which is a quadratic in $b_3.$ We know that $b_3 = -b_1-b_2$ is a solution to this quadratic since $h(x) = -f(x) - g(x)$ works, so since the sum of the roots is $-2b_1 - 2b_2$ by Vieta, we have that $b_3 = -b_1-b_2$ is the unique solution. Now, plugging into any equation lets us solve for $c_3$, and so hence $h(x) = -f(x) - g(x)$ is uniquely determined in this case. Case 2. $a_1 + a_2 + a_3 \neq 0$, and $\left < a_2, c_2 \right >$ and $\left < a_1, c_1 \right >$ are linearly independent Our three equations here read that $$4a_2c_3 + 4c_2a_3 = 2b_2b_3 + 1,$$$$4a_1c_3 + 4c_1a_3 = 2b_1b_3 + 1,$$$$b_3^2 = 4a_3c_3+1,$$where $b_3 = -b_1-b_2$ is a constant now. Since $\left < a_2, c_2 \right >$ and $\left < a_1, c_1 \right >$ are linearly independent, the first two equations uniquely determine $a_3, c_3$, so therefore we must have $h(x) = -f(x) - g(x)$ as the unique solution. Case 3. $\left < a_2, c_2 \right >$ and $\left < a_1, c_1 \right >$ are linearly dependent Our three equations here still read that $$4a_2c_3 + 4c_2a_3 = 2b_2b_3 + 1,$$$$4a_1c_3 + 4c_1a_3 = 2b_1b_3 + 1,$$$$b_3^2 = 4a_3c_3+1,$$where $b_3 = -b_1-b_2$ is still a constant. Observe that if $c_1 = 0$, then $c_2 = 0$ as well, and vice versa. If both $c_i$'s are zero, then we have that $b_1^2 = b_2^2 = 1$, so that $0 = 4a_1c_2 + 4a_2c_1 = 2b_1b_2 + 1 = \pm 1,$ contradiction. Hence, we've that $c_1, c_2$ are both nonzero. Therefore, we have that $$2b_1b_2+1 = 4a_1c_2 + 4a_2c_1 = \pm 2 \sqrt{(4a_1c_1) \cdot (4a_2c_2)} = \pm 2 \sqrt{(b_1^2 - 1)(b_2^2-1)},$$where we used the fact that $a_1c_2 = a_2c_1 = \pm \sqrt{a_1c_1a_2c_2}.$ Squaring both sides and simplifying then gives us that $$4(b_1^2 + b_1b_2 + b_2^2) = 3.$$Let $r_1 = 4a_2c_3, r_2 = 4c_2a_3.$ Notice that $r_1, r_2$ are roots of the quadratic polynomial $$x^2 - (2b_2b_3+1)x + (b_3^2-1)(4a_2c_2) = 0.$$Note that all coefficients of the quadratic above are constant. We claim that this quadratic is actually a square, which would hence fix the value of $r_1 = r_2$, hence allowing us to find $a_3 = \frac{r_2}{4c_2}$ and $c_3 = \frac{r_1}{4a_2}$ uniquely. This would solve the problem since those unique values must be the values when $h(x) = -f(x) - g(x).$ So let's show that this quadratic is a square. Indeed, observe that its discriminant is $$(2b_2b_3+1)^2 - 16a_2c_2(b_3^2-1) = (2b_2(-b_1-b_2)+1)^2 - (4b_2^2 - 4)((b_1+b_2)^2-1),$$which simplifies to $$4(b_1^2 + b_1b_2 + b_2^2) - 3 = 0,$$as desired. $\square$

\begin{align*} a(c+f+i) + c(a+d+g) &= 2b(b+e+h) \\ d(c+f+i) + f(a+d+g) &= 2e(b+e+h) \\ g(c+f+i) + i(a+d+g) &= 2h(b+e+h) \end{align*}$\blacksquare$

Since the three crucial algebraic manipulations have been highlighted with a $\blacksquare$, I'll explain the process of coming up with those in the first place. $\color{green} \rule{25cm}{2pt}$ $\color{red} \text{Sec 1: Null constants.}$ After I saw the problem and tested what would happen if $f+g+h \equiv 0$ (that is, as @above pointed, would be equivalent to $f,g,f+g$ having discriminant $1$), I thought that this must be a play on coefficients instead of roots (though switching back and forth between the two brought a lot of insights as to where to go). After not wanting to set $a_i,b_i,c_i$ as variables with $1 \leq i \leq 3$, I thought I'd give individual letters as (personalities) of the variables. First idea was the standard manipulation of $(b+e)^2 - 4(ac+af+cd+df) = 1$, which after cancelling out $b^2 = 4ac+1$ and the like yields \[ be - 2af - 2cd = \dfrac{1}{2} \]which wasn't $\textit{really}$ a bad thing, but I had no other good plans with this, so I tried to route this back into $be = 2af+2cd+\dfrac{1}{2}$ and square/replace $b^2,e^2$ with more $a,c,d,f$ but this prompted a really bad-looking equation. That got me the idea of factoring out $4(af-cd)^2$, though. Another idea which resembled this was considering $r_1,r_2$ of the equation $ax^2+bx+c$, but this immediately broke up the nicely-normalized six equations into \[ r_1,r_2 = \dfrac{-b\pm1}{2a} \]which (almost) cannot be homogenized back, as we only have $1$ as a sum of degree-two terms; when we need $1$ to be the same degree as $-b$ (i.e. of degree $1$) to be able to interact with the six equations. By then, since there doesn't seem to be a way to simplify the nine variables $a$ to $i$ (didn't consider Vieta jumping as a thing here), I opted for the tabular representation \[\begin{tabular}{ccc|} $a$ & $b$ & $c$ \\ \hline $d$ & $e$ & $f$ \\ \hline $g$ & $h$ & $i$ \end{tabular} \\ \]and wrote the equation $b^2-4ac = 1$ on the first row, et cetera, and $b^2+2be+e^2-4ac-4af-4cd-4df = 1$, et cetera, to the right of the $``\text{hline}"$s. As I noticed that all the equations of $b,e,h$ are wrapped in squares, as we're given the terms \[ b^2,e^2,h^2,(b+e)^2,(e+h)^2,(h+b)^2 = \ \text{something} \]I thought it might be good to create $(b+e+h)^2$, and so that's what I did first: adding the equations of form $(b+e)^2$ and substracting it with the sum $b^2+e^2+h^2$ to yield \[ (b+e+h)^2 = 4 (a+d+g)(c+f+i) \]and another thought came up: if $b,e,h$ are indeed special (and $a,c,d$,etc are common), why is $4$ always stalking those common terms? Moreover, I thought that the constant $1$ is also redundant, we might as well replace $b+e+h$ into whatever we want (and this process is exactly what homogenisation is about). So, to remove the hindrance of $b^2 = 4ac+1$ and the like, it's easily handled by substituting $a' = 2a$ and let $a'$ to be the new $a$. That will transform $f(x) = ax^2+bx+c$ into $\dfrac{a'}{2}x^2+bx+\dfrac{c'}{2}$. $\color{green} \rule{25cm}{0.5pt}$ With this simplified form, I gained the courage to write down $b^2-1=ac$ and analyze the variables $\textit{independently}$ (instead of over-relying on $(b+e+h)^2$, for instance) --- but I was $\textsf{dismayed}$ to see that there's no satisfactory way to solve the $\textbf{original assertion}$ in two variables. In short, the equations $ac = b^2-1, df =e^2-1, af+cd = 2be+1$ had no good general form, so I was convinced that while I should analyze the variables independently, their relation with each other are more intricate than I initially expected (I assumed that analyzing $b+e+h$ and forcing it to be 0 in some way would be enough to guarantee $f+g+h \equiv 0$ easily, but I realized the fact that no simple solutions exist means the problem is hard.) Still, the new six equations given at the second paragraph of the Solution is much better to look at --- so that's what I was using till the end of my working. (Note that we haven't even reached the first checkpoint here.) $\color{green} \rule{25cm}{2pt}$ $\color{red} \text{Sec 2: Three-term equations + (Zero) issues with three-term variables.}$ Since almost all hope of drawing $f(x),g(x),h(x)$ into diagrams is gone, as $D = 1$ is just too strong a statement to be represented as graphs with no information loss, I kept my eyes strictly at the $b,e$ and $h$. Interestingly, before getting the strange idea to act in the first checkpoint, simply doing the obvious AM-GM on \[ b^2+e^2 = ac+df+2 \geq 2be = af+cd-1 \Rightarrow (a-d)(c-f) \geq -3 \]won't affect much; we'll only know that $(a-d)(c-f)$ cannot be hugely negative --- and the term $-3$ is also a fake constant; since the $-1$s that exist only serves as the discriminant equalizer. After this, (still) banking on the relations between $b^2,be$ and other stuff, I had the idea to factor $b+e+h$ and let the LHS speak for itself, and it turned into the miraculous equation(s) on the first checkpoint. From here, I tried to solve $c+f+i$, $a+d+g$ and $b+e+h$ in a similar manner to @above's post, but it just turned out to be a calamity, as it would just root back to the term \[ afh-aei+bdi+cdh-cge+bfg = 0\]or something similar (since asking to solve a random three-variable equation with coefficients $\textit{depending}$ on the variables $\textit{being solved for}$ is not a good idea.) However, the prevalence of $a+d+g$ tempted me heavily to consider the term \[ (a+d+g)(c+f+i) = (b+e+h)^2\]as a constant; I let $(b+e+h)^2 = N^2$ for that manner. This certainly motivated me to cover the miraculous equations into \[ a(c+f+i)+c(a+d+g) = 2b \sqrt{(c+f+i)(a+d+g)} \]while crossing the $\pm$ on the RHS for now. Again, this motivated me further to consider $c+f+i$ and $a+d+g$ independently as $X^2$ and $Y^2$ (assuming they are indeed positive --- I was turning a blind eye on the sums possibly being negative (as I thought I could force them to be the opposite sign in some WLOG argument). This turned out nicer than expected with results such as \[ \dfrac{b\pm1}{a} = \dfrac{e\pm1}{d} = \dfrac{h\pm1}{f} \]but this was still not definite. So, the final idea before reaching my conclusion was to remove the signs entirely by squaring them and factoring, which unexpectedly yields the second and third checkpoint! \[\begin{tabular}{p{12cm}} $\rule{4cm}{0.5pt} \hspace{1cm} \blacksquare \hspace{1cm} \rule{4cm}{0.5pt}$ \\ \end{tabular} \\ \]