<s>[INST] Find all real solutions of $\sqrt{x} - \sqrt{2-2x} = 1$ Squaring both sides of $\sqrt{x} - \sqrt{2-2x} = 1$ and rearranging 
I arrive at the quadratic $9x^2 - 10x + 1 = 0$ which has solutions $x=1/9$ and $x=1$.
I don't understand why $x=1$ fits the original equation but $x=1/9$ doesn't (left hand side gives $-1$).
 [/INST] $$\sqrt{x} - 1 = \sqrt{2-2x}$$
$$x -2\sqrt{x} + 1 = 2-2x$$
$$3x -2\sqrt{x} - 1 = 0$$
Now if $\sqrt{x} = u$ we have:
$$3u^2 -2u - 1 = 0$$
$$u = 1 \vee u=-\frac{1}{3}$$
In the case $u = 1$ we've found the solution $x = 1$. But in the case $u = -\dfrac{1}{3}$ we get:
$$\sqrt{x} = -\frac{1}{3}$$
Do you see the issue now?
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