Find all real solutions $x$ for the equation $x^{1/2} − (2−2x)^{1/2} = 1$

Question

This is what the answer says:

Note that the equation can be rewritten as $\sqrt{x} − \sqrt{2 − 2x} = 1$,

and the existence of such real $x$ implies that $x$ is larger than or equal to $0$ and $x$ is less than or equal to $1$, since we implicitly assume that $\sqrt{x}$ and $\sqrt{2 − 2x}$ are real as well. Then

$\sqrt{x} − \sqrt{2 − 2x} < \sqrt{1} = 1 \text{ if } x < 1$, and

$\sqrt{x} − \sqrt{2 − 2x} = 1 \text{ if } x = 1$.

Thus, $x = 1$ is the only real solutions of the equation $\sqrt{x} − \sqrt{2 − 2x} = 1$.

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I don't understand how

$\sqrt{x} − \sqrt{2 − 2x} < \sqrt{1} = 1 \text{ if } x < 1$.

I understand everything else.

Can someone please explain? Thank you.

Answer 1

Notice that $\sqrt{x}-\sqrt{2-2x} \leq \sqrt{x}$. Then if $x<1$ we have $\sqrt{x}<\sqrt{1}=1$. It follows that $\sqrt{x}-\sqrt{2-2x} <1$ for $x<1$. So if $x<1$ it is not a solution. If $x=1$ we have $\sqrt{x}-\sqrt{2-2x}=\sqrt{1}-\sqrt{2-2}=1 $. So $x=1$ is a solution

Answer 2

Observe that $$\sqrt{x} - \sqrt{2-2x} = 1 \implies x = 1 + 2-2x + 2\sqrt{2-2x} \implies 3x - 3 = \sqrt{2-2x}$$ $$ \implies 9(x^2 - 2x + 1) = 2-2x \implies 9x^2 - 16x +7 = 0 \implies (x -1)(9x - 7) = 0 $$ Thus $ x = 1, \frac{7}{9} $. Now, $$ \sqrt{\frac{7}{9}}-\sqrt{2-\frac{14}{9}} \neq 1 $$ Thus $x=1 $ is the solution.