A Problem From The Future!

$\begin{aligned} \sqrt{2027x^2+3047x-2030} - \sqrt{2026x^2+3047x-2005} & = x-5 \\ \sqrt{{\color{#3D99F6}2026x^2+3047x-2005} + x^2 - 25} - \sqrt{\color{#3D99F6}2026x^2+3047x-2005} & = x-5 \end{aligned}$

Let $u = 2026x^2+3047x-2005$ , then

$\begin{aligned} \sqrt{{\color{#3D99F6}u} + x^2 - 25} - \sqrt{\color{#3D99F6}u} & = x-5 & \small \color{#3D99F6} \text{Rearranging} \\ \sqrt{u+x^2-25} & = x-5 + \sqrt u & \small \color{#3D99F6} \text{Squaring both sides} \\ u + x^2 - 25 & = x^2 - 10x + 25 + 2(x-5)\sqrt u + u \\ 2(x-5)\sqrt u - 10x + 50 & = 0 \\ \implies (x-5)(\sqrt u - 5) & = 0 \end{aligned}$

$\implies \color{#3D99F6} x = 5$ is a root and

$\begin{aligned} \sqrt u & = 5 & \small \color{#3D99F6} \text{Squaring both sides} \\ 2026x^2+3047x-2005 & = 25 & \small \color{#3D99F6} \text{As }u = 2026x^2+3047x-2005 \\ 2026x^2+3047x-2030 & = 0 \\ (2x - 1)(1013x +2030) & = 0 \end{aligned}$

$\implies x = \begin{cases} \color{#3D99F6} \frac 12 \\ \color{#D61F06}- \frac {2030}{1013} & \small \color{#D61F06} \text{Not a root when check with the original equation.} \end{cases}$

Therefore the roots are $\frac 12$ and $\frac 51$ and $|ab-cd| = |1\times 2 - 5 \times 1| = \boxed 3$ .

Let $\sqrt{2027x^{2}+3047x-2030}=A$ and $\sqrt{2026x^{2}+3047x-2005}=B$ , then the question says, $A-B=x-5$ .

Incidentally, $A^{2}-B^{2}=x^{2}-25$ . Dividing these two equations we get either $\boxed{x=5}$ or $A+B=x+5$ .

Furthermore, solving these linear equations, we have, $A=x$ and $B=5$ .

Substituting back in the above equation, we have, $\sqrt{2027x^{2}+3047x-2030}=x \Rightarrow 2026x^{2}+3047x-2030=0$ .

The equation for $B$ also gives the same quadratic equation.

Solving this quadratic equation we get $\boxed{x=\frac{1}{2}}$ and $x=-\frac{2030}{1013}$ .

We neglect the negative answer as the output of a square root cannot be negative.

Thus the two roots are $x=5$ and $x=\frac{1}{2}$ which gives the final answer as $|1\times2 - 5\times1| = 3$ .