How to solve system of equations involving square roots

Question

How to solve the following system of equations? I've tried some basic techniques like adding/substracting and squaring but with no effect. $$ \left\{ \begin{array}{c} \sqrt{1 + x_1} + \sqrt{1 + x_2} + \sqrt{1 + x_3} + \sqrt{1 + x_4} = 2\sqrt{5} \\ \sqrt{1 - x_1} + \sqrt{1 - x_2} + \sqrt{1 - x_3} + \sqrt{1 - x_4} = 2\sqrt{3} \end{array} \right. $$

Answer 1

Let $\mathbf{r}_k= \begin{pmatrix} \sqrt{1+x_{k}} \\ \sqrt{1-x_{k}} \end{pmatrix}$, then $\displaystyle \sum_{k=1}^{4} \mathbf{r}_k= \begin{pmatrix} 2\sqrt{5} \\ 2\sqrt{3} \end{pmatrix}$

Now, \begin{align*} \left| \sum_{k=1}^{4} \mathbf{r}_k \right| &=2\sqrt{5+3} \\ &= 4\sqrt{2} \\ \sum_{k=1}^{4} |\mathbf{r}_k| &= \sum_{k=1}^{4} \sqrt{2} \\ &= 4\sqrt{2} \end{align*}

Considering the inequality $$\left| \sum_{k=1}^{4} \mathbf{r}_k \right| \le \sum_{k=1}^{4} |\mathbf{r}_k|$$

in which equality holds if and only if all $\mathbf{r}_{k}$ are equal.

That is $$4 \begin{pmatrix} \sqrt{1+x_k} \\ \sqrt{1-x_k} \end{pmatrix}= \begin{pmatrix} 2\sqrt{5} \\ 2\sqrt{3} \end{pmatrix}$$

Hence $$\fbox{$x_1=x_2=x_3=x_4=\frac{1}{4}$}$$

Answer 2

HINT:-

Since the roots have to be positive, we see that each radical must reduce to the form $a\sqrt5$ for the first equation , and $b\sqrt3$ for the second equation.