Solving $\sqrt{1+\sqrt{2x-x^2}} + \sqrt{1-\sqrt{2x-x^2}} = \sqrt{4-2x}$

Question

Can we find the solutions for this equation? $$\sqrt{1+\sqrt{2x-x^2}} + \sqrt{1-\sqrt{2x-x^2}} = \sqrt{4-2x}, \quad x \in \mathbb{R}$$ I tried to amplify the second square root in the $LHS$ with the conjugate and then use AM-GM in order to find where $x$ can be.

Also, the existence conditions imply $x \leq 2$. I obtained $x \leq \frac{4}{3}$.

Answer 1

Note

$$LHS^2= \left(\sqrt{1+\sqrt{2x-x^2}} + \sqrt{1-\sqrt{2x-x^2}} \right)^2\\ =2+2\sqrt{(1 -x)^2}= {4-2x}=RHS^2 $$

So, the solution is just the domain of the equation, which is jointly determined by

$$2x-x^2\ge 0,\>\>\>1-\sqrt{2x-x^2}\ge 0,\>\>\>4-2x\ge 0$$ which leads to $0\le x\le 1$.

Answer 2

The domain gives $0\leq x\leq2$.

Now, $$\sqrt{1+\sqrt{2x-x^2}} + \sqrt{1-\sqrt{2x-x^2}} = \sqrt{\left(\sqrt{1+\sqrt{2x-x^2}} + \sqrt{1-\sqrt{2x-x^2}}\right)^2}=$$ $$=\sqrt{2+2\sqrt{1-2x+x^2}}=\sqrt{2+2|x-1|}.$$ Thus, it's enough to solve $$1+|x-1|=2-x.$$ Can you end it now?

I got $0\leq x\leq1$.

Answer 3

\begin{align*} &\Rightarrow\sqrt{4-2x}\\ &=\sqrt{2+2(1-x)}&(x\in[0,2])\\ &=\sqrt{[1-\sqrt{2x-x^2}]+[1+\sqrt{2x-x^2}]+2\color{red}{(1-x)}}\\ &=\sqrt{[1-\sqrt{2x-x^2}]+[1+\sqrt{2x-x^2}]+2\color{red}{\sqrt{(1-x)^2}}}&(\color{red}{x\in[0,1]})\\ &=\sqrt{[1-\sqrt{2x-x^2}]+[1+\sqrt{2x-x^2}]+2\sqrt{1-(2x-x^2)}}\\ &=\sqrt{[1-\sqrt{2x-x^2}]+[1+\sqrt{2x-x^2}]+2\sqrt{1-\sqrt{2x-x^2}}\sqrt{1+\sqrt{2x-x^2}}}\\ &=\sqrt{\left(\sqrt{1-\sqrt{2x-x^2}}+\sqrt{1+\sqrt{2x-x^2}}\right)^2}\\ &=\sqrt{1-\sqrt{2x-x^2}}+\sqrt{1+\sqrt{2x-x^2}} \end{align*}

Answer 4

You may first symmetrise $x(2-x)$ by setting $\boxed{x=y+1}$:

$$\sqrt{1+\sqrt{1-y^2}} + \sqrt{1-\sqrt{1-y^2}}= \sqrt{2(1-y)} \text{ with } -1\leq y\leq 1$$

Now, squaring gives

$$2+2|y|= 2(1-y) \Leftrightarrow |y| = -y$$

Hence, with $-1\leq y\leq 1 \stackrel{|y|=-y}{\Rightarrow} -1\leq y\leq 0 \stackrel{x=y+1}{\Rightarrow} \boxed{0 \leq x \leq 1}$ is the solution.

Answer 5

The square roots are all well defined real numbers if and only if $0\le x\le2$. In this range, the numbers on both sides are positive, so we can say that

$$\begin{align} LHS=RHS &\iff\left(\sqrt{1+\sqrt{2x-x^2}}+\sqrt{1-\sqrt{2x-x^2}}\right)^2=\left(\sqrt{4-2x}\right)^2\\ &\iff2+2\sqrt{1-(2x-x^2)}=4-2x\\ &\iff1+\sqrt{(1-x)^2}=2-x\\ &\iff|1-x|=1-x\quad\text{(note the absolute values!)}\\ &\iff1-x\ge0\\ &\iff x\le1 \end{align}$$

Together with the restriction $0\le x\le2$, we have

$$\sqrt{1+\sqrt{2x-x^2}}+\sqrt{1-\sqrt{2x-x^2}}=\sqrt{4-2x}\iff 0\le x\le1$$

Remark: If it seems odd that the identity suddenly stops working at $x=1$, note that what we really have is the two-piece identity

$$\sqrt{4-2x}= \begin{cases} \sqrt{1+\sqrt{2x-x^2}}+\sqrt{1-\sqrt{2x-x^2}}\quad\text{for }0\le x\le1\\ \sqrt{1+\sqrt{2x-x^2}}-\sqrt{1-\sqrt{2x-x^2}}\quad\text{for }1\le x\le2 \end{cases}$$