Find all $z$ such that $\sqrt{5z+5} - \sqrt{3 - 3z} - 2\sqrt{z} = 0$. Find all $z$ such that $\sqrt{5z+5} - \sqrt{3 - 3z} - 2\sqrt{z} = 0$.
After much trial and error, I was able to rearrange the above equation
as $80z^2 - 112z-4 = 0$.
$z = \frac{7\pm3\sqrt{6}}{10}$.
I substituted $\frac{7+3\sqrt{6}}{10}$ as $z$ in the $\sqrt{5z+5} - \sqrt{3 - 3z} - 2\sqrt{z} = 0$. Things worked out.
Am I still not done? Must I check for extraneous solutions in this situation?
I wonder if it might not require it because if ever we know one of the roots of a quadratic with rational coefficients is irrational, we know the other.
I suppose I'm also trying to avoid going through another round of substitution and test with these irrational conjugates. I'm not very trusting of my algebra yet. So I end up rewriting the same equations time and time again.
 A: In order to have a clean complete solution, the "steps in between" have to be mentioned, and moreover it should be clear if we work with $\Rightarrow$ (only), or if we work with equivalences all the time.
For instance, if we take the initial equation, and move some terms "on the other side", then take squares, then we are going into one direction (only). From the given equation we obtain the new equation, (i.e. the given equation implies the new equation), but not necessarily in the converse direction also. With such a strategy, one must finally verify the obtained solutions. (Or do something instead.)
Best, we apply squaring on quantities that are both positive. Then squaring leads to a new equation, but we have an equivalence, a particular value $z_0$ is solution of the first equation if and only if it is a solution of the second equation.
In this sense, we can deliver the solution as follows (so that no final check is needed).


*

*Let $z$ be a solution of the equation $$\sqrt{5z+5} - \sqrt{3 - 3z} - 2\sqrt{z} = 0\ ,$$ where all square roots are taken from values $\ge 0$. (And the results are $\ge 0$.)


*(In other words, $5z+5\ge 0$, and $3-3z\ge 0$, and $z\ge 0$. Equivalently, $0\le z\le 1$. And it is the duty of the person writing down the problem to insert this condition in the problem. But well, teachers often mention "so that the equation makes sense". And more often they tacitly (i.e. do not) mention it. Well, for me, the stuff makes sense also for $z=1+i$, and i will take my own branch of the square root to have more solutions. Long story short: Didactically, the problem must come with $0\le z\le 1$. So...)


*We restrict to values of  $z$ (and thus also to solutions $z$)
with $0\le z\le 1$. (The only values, that "make sense in the school" while writing the equation. (Well, only in some very special schools respecting the tradition of old centuries, i.e. those schools in those countries where students, parents, teachers, and the piece with educational responsibility in the government insist to live in the older centuries. Such schools deny the knowledge of $i$, and insist to put all the responsibility for the claims and proofs on the student=solver.))


*Now we rewrite equivalently at each step, i.e. from line to next line:
$$
\begin{aligned}
\sqrt{5z+5} - \sqrt{3 - 3z} - 2\sqrt{z} &= 0\ ,\\
\sqrt{5z+5} &= \sqrt{3 - 3z} + 2\sqrt{z} \ ,\\
&\qquad\text{(squaring quantities $\ge 0$)}\\
5z+5 &= 3-3z + 4z +4\sqrt{3z(1-z)}\ ,\\
4z+2 &= 4\sqrt{3z(1-z)}\ ,\\
2z+1 &= 2\sqrt{3z(1-z)}\ ,\\
&\qquad\text{(squaring quantities $\ge 0$)}\\
4z^2+4z+1 &= 12z(1-z)\ ,\\
4z^2+4z+1 &= 12z-12z^2\ ,\\
16z^2-8z+1 &= 0\ ,\\
(4z-1)^2 &= 0\ ,\\
(4z-1) &= 0\ ,\\
4z&=1\ ,\\
z&=\frac 14\ .
\end{aligned}
$$
In this moment, we know that $1/4\in[0,1]$, makes sense for the given equation, is a solution, and the only solution. (There is no need to check it.)
Note: In this special case it is easier to work in only one direction, using implications $\Rightarrow$, and finally verify this solution:
$$
\underbrace{\sqrt{5\cdot \frac 14+5}}_{=\frac 52}
-
\underbrace{\sqrt{3-3\cdot \frac 14}}_{=\frac 32}
-
2\sqrt{\frac 14}
=0\ .
$$
(Yes, $\frac 52-\frac 32-1=0$.)
A: $$\sqrt{5 z+5}=2 \sqrt{z}+\sqrt{3-3 z}\tag{1}$$
Square both sides
$$5z+5=4z+3-3z+4\sqrt{3z-3z^2}$$
$$2+4z=4\sqrt{3z-3z^2}\to 1+2z=2\sqrt{3z-3z^2}$$
Square again both sides
$$1+4z+4z^2=12z-12z^2$$
$$16 z^2-8 z+1=0\to (4z-1)^2=0\to z=\frac14$$
Which is actually a solution of $(1)$ (verify!)
