$3x^2-7xy-6y^2-2x+17y-5=0$
My original goal here was to know whether or not this was a degenerate conic, so I isolated $x$ by applying the Quadratic Formula.
$3x^2+(-7y-2)x+(-6y^2+17y-5)=0$
$x = \frac{-(-7y-2)\pm\sqrt{(-7y-2)^2-4(3)(-6y^2+17y-5)}}{2(3)}$
a) $x = \frac{(7y+2)\pm\sqrt{(11y-8)^2}}{6}$
It is to my understanding that an expression of the form $\sqrt{x^2}$ is equal to $|x|$ for every $x$ $\in$ $\Bbb{R}$, which is why I expect a to become
$x = \frac{(7y+2)\pm|11y-8|}{6}$
and so yielding two equations
b)$$x = \frac{(7y+2)+|11y-8|}{6}; x = \frac{(7y+2)-|11y-8|}{6}$$
However, my book does not include the absolute values, $\sqrt{(11y-8)^2}$ simply becomes $11y-8$
giving
c)$$x = \frac{3y-1}{6}; x = \frac{-2y+5}{3}$$
Moreover, each equation in b and c have different graphs, but coincide when combined. What am I missing? Why do we get rid of the absolute value?
I think this is because of the $\pm$ in the Quadratic Formula. My computations are as follows:
$$x = \frac{(7y+2)\pm|11y-8|}{6}$$
By definition of absolute value,
Case 1 $+$
$$x = \frac{(7y+2)+|11y-8|}{6}$$
Let $11y-8\geq 0$ $\implies y\geq\frac{8}{11}$ $\implies x\geq\frac{13}{11}$ ,
$$x = \frac{7y+2+11y-8}{6}$$ $$x = 3y-1$$ $$y = \frac{x+1}{3}$$ $$\frac{x+1}{3}\geq\frac{8}{11}$$ $$x\geq\frac{13}{11}$$
Let $11y-8\lt0$ $\implies y\lt\frac{8}{11}$ $\implies x\gt\frac{13}{11}$,
$$x = \frac{7y+2-11y+8}{6}$$ $$x = \frac{-4y+10}{6}$$ $$x = \frac{-2y+5}{3}$$ $$y = \frac{5-3x}{2}$$ $$\frac{5-3x}{2}\lt\frac{8}{11}$$ $$x\gt\frac{13}{11}$$
This means we can write $x = \frac{(7y+2)+|11y-8|}{6}$ as the combination of $y = \frac{x+1}{3}$ and $y = \frac{5-3x}{2}$, with $x\geq\frac{13}{11}$ and $y \in \Bbb{R}$ because $y\geq\frac{8}{11}$ or $y\lt\frac{8}{11}$.
The graph looks like this:
Case 2 $-$
$$x = \frac{(7y+2)-|11y-8|}{6}$$
Let $11y-8\geq 0$ $\implies y\geq\frac{8}{11}$ $\implies x\leq\frac{13}{11}$,
$$x = \frac{7y+2-11y+8}{6}$$ $$x = \frac{-4y+10}{6}$$ $$x = \frac{-2y+5}{3}$$ $$y = \frac{5-3x}{2}$$ $$\frac{5-3x}{2}\geq\frac{8}{11}$$ $$x\leq\frac{13}{11}$$
Let $11y-8\lt0\implies y\lt\frac{8}{11} \implies x\lt\frac{13}{11}$,
$$x = \frac{7y+2+11y-8}{6}$$ $$x = 3y-1$$ $$y = \frac{x+1}{3}$$ $$\frac{x+1}{3}\lt\frac{8}{11}$$ $$x\lt\frac{13}{11}$$
This means we can write $x = \frac{(7y+2)-|11y-8|}{6}$ as the combination of $y = \frac{x+1}{3}$ and $y = \frac{5-3x}{2}$, with $x\leq\frac{13}{11}$ and $y \in \Bbb{R}$.
The graph looks like this:
This means we can write $x = \frac{(7y+2)\pm|11y-8|}{6}$ as the combination of $y = \frac{x+1}{3}$ and $y = \frac{5-3x}{2}$, with $x \in \Bbb{R}$ because $x\geq\frac{13}{11}$ or $x\leq\frac{13}{11}$, and $y \in \Bbb{R}$.
In other words,
We can think of $x = \frac{(7y+2)\pm\sqrt{(11y-8)^2}}{6}$, which is $x = \frac{(7y+2)\pm|11y-8|}{6}$
as $x = \frac{(7y+2)+|11y-8|}{6}$ and $x = \frac{(7y+2)-|11y-8|}{6}$
or as $y = \frac{x+1}{3}$ (purple) and $y = \frac{5-3x}{2}$ (orange)
Note: I used GeoGebra for the graphs.
If my analysis is correct, is there any reputable sources (or proof) that says we can simply cancel out under these circumstances?
$$\pm\sqrt{(11y-8)^2} = \pm(11y-8)$$
Your question essentially reduces to $\pm |x| = \pm x$. Note that $\pm x$ isn't really a value but rather the set $\{x, -x\}$, so we want to show $\{|x|, -|x|\} = \{x, -x\}$. This is simple casework: