# How to solve system of equations involving square roots

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

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

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.

• Could you please tell something more? I have no idea how to use this observation. – XiaoHensZimmer Dec 9 '16 at 16:31
• The $x_i$ are not necessarily rational, so this hint does not apply. – Théophile Dec 9 '16 at 19:55

Denote $$\sqrt{1+x_i}=a_i,~~~\sqrt{1-x_i}=b_i, ~~~(i=1,2,3,4)$$

It's clear that $$a_i,b_i \geq 0,$$and$$a_i^2+b_i^2=2.\tag1$$

Moreover, the system of equations could be rewritten as $$\sum_{i=1}^4a_i=2\sqrt{5},~~~~\sum_{i=1}^4b_i=2\sqrt{3}.\tag2$$

From $(1)$, we have $$\sum_{i=1}^4a_i^2+\sum_{i=1}^4b_i^2=8.$$

From $(2)$, we have $$\left(\frac{1}{4}\sum_{i=1}^4a_i\right)^2+\left(\frac{1}{4}\sum_{i=1}^4b_i\right)^2=2.\tag3$$

But by the inequality $A_n \leq Q_n$, we have

$$\left(\frac{1}{4}\sum_{i=1}^4a_i\right)^2+\left(\frac{1}{4}\sum_{i=1}^4b_i\right)^2 \leq \frac{1}{4}\sum_{i=1}^4a_i^2+\frac{1}{4}\sum_{i=1}^4b_i^2=\frac{1}{4}\left(\sum_{i=1}^4a_i^2+\sum_{i=1}^4b_i^2\right)=2.\tag4$$

Form $(3)$, we see that the equality holds in $(4)$. Recall the condition for equality in $A_n \leq Q_n$, which is $$a_1=a_2=a_3=a_4,~~~b_1=b_2=b_3=b_4.$$

Thus, we may obtain $$a_i=\frac{\sqrt{5}}{2},~~~b_i=\frac{\sqrt{3}}{2}.$$

As a result, $$x_i \equiv\frac{1}{4},$$ where $i=1,2,3,4.$