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Solve for $x \in \mathbb{R}$

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

I tried some substitutions and squaring but that didn't help. I also tried to use inequalities as done in my previous problem, but that too didn't help.

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  • $\begingroup$ I don't really see why this would have an algebraic solution. You are going to end up mixing square roots with $x\sqrt{x}$. $x$ and $\sqrt{x}$ is fine since that should be transformable to a quadratic but not if you have $\sqrt{x}$ and $x\sqrt{x}$. It seems this would be akin to (possibly) a cubic (but not even sure about that). $\endgroup$ – Jared May 4 '16 at 2:04
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Given : $ 1 + \dfrac{\sqrt{x+3}}{1+\sqrt{1-x}} = x + \dfrac{\sqrt{2x+2}} {1+\sqrt{2-2x}} $

Let $\alpha = x+3 $ ; $\beta = 2x+2$ ; $\left(\beta - \alpha\right) = x-1$

$\implies 1+ \dfrac{\sqrt{\alpha}}{1+\sqrt{4-\alpha}} = x + \dfrac{\sqrt{\beta}} {1+\sqrt{4-\beta}} $

$\implies \alpha + \dfrac{\sqrt{\alpha}}{1+\sqrt{4-\alpha}} = \beta + \dfrac{\sqrt{\beta}} {1+\sqrt{4-\beta}} $

Let $f(x) = x + \dfrac{\sqrt{x}}{1+\sqrt{4-x}}$, the given equation becomes,

$f(\alpha) = f(\beta)$

Note that $f(x)$ is monotonic increasing in its domain,

$\therefore \alpha = \beta$

$\implies x+3 = 2x +2$

$\implies x=\boxed{1}$

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  • $\begingroup$ Hello Sir. Can you also help me with this question ? $\endgroup$ – Henry May 4 '16 at 5:35
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I hope that my solution is simpler: \begin{eqnarray} &&1+\frac{\sqrt{x+3}}{1+\sqrt{1-x}}=x+\frac{\sqrt{2x+2}}{1+\sqrt{2-2x}}\\ &\Longleftrightarrow& x-1+\frac{\sqrt{2x+2}}{1+\sqrt{2-2x}}-\frac{\sqrt{x+3}} {1+\sqrt{1-x}}=0\\ &\Longleftrightarrow& x-1+\frac{\sqrt{2x+2}+\sqrt{2x+2}\sqrt{1-x}-\sqrt{x+3}-\sqrt{x+3}\sqrt{2-2x}}{(1+\sqrt{2-2x})(1+\sqrt{1-x})}=0\\ &\Longleftrightarrow& x-1+\frac{\frac{x-1}{\sqrt{2x+2}+\sqrt{x+3}}+\frac{4(x-1)}{\sqrt{2x+2}\sqrt{1-x}+\sqrt{x+3}\sqrt{2-2x}}}{(1+\sqrt{2-2x})(1+\sqrt{1-x})}=0\\ &\Longleftrightarrow& (x-1)\left(1+\frac{\frac{1}{\sqrt{2x+2}+\sqrt{x+3}}+\frac{4}{\sqrt{2x+2}\sqrt{1-x}+\sqrt{x+3}\sqrt{2-2x}}}{(1+\sqrt{2-2x})(1+\sqrt{1-x})}\right)=0\\ &\Longleftrightarrow& x=1. \end{eqnarray}

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