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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 $-1$).

user7875067
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    Because when one replaces $u=1$ by $u^2=1$ one replaces the set of $u$-solutions ${1}$ by ${1,-1}$. In your case you replaced the set of $x$-solutions ${1}$ by ${1,\frac19}$. – Did Oct 20 '17 at 17:21
  • Always remember $a=b\iff a^2=b^2$ is false ! –  Oct 20 '17 at 17:28

3 Answers3

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$$\sqrt{x} - 1 = \sqrt{2-2x}$$ $$x -2\sqrt{x} + 1 = 2-2x$$ $$3x -2\sqrt{x} - 1 = 0$$

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?

orlp
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Just like dividing by $f(x)$ loose you the solution $x_0$ where $f(x_0)=0$ Multiply by $f(x)$ can add you solutions, so you need to check if the solutions you got are actually working.

ℋolo
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Let $x=u^2$, so that $\sqrt x=u$ and $u\ge0$.

Then $$\sqrt{2-2u^2}=u-1$$

which implies $u\ge1$, which implies $2-2u^2\le0$.

We needn't go further, the only solution is $u=1$.