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.

  • $\begingroup$ 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

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,

$\therefore \alpha = \beta$

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

$\implies x=\boxed{1}$

| cite | improve this answer | |

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}

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.