1. Solve Equation

Hii !

Can You Help Me For This Equation ,

Solve in IR the Equation :

$x=a-\sqrt{a-\sqrt{x}}$

following the values of parameter a ( $a \in \mathbb{R^+}$ )

2. Well

What have you tried so far? Have you subtracting an a from both sides, and then square both sides?

3. $x= a-\sqrt{a-\sqrt{x}}$
$\Longleftrightarrow a-x=\sqrt{a-\sqrt{x}}$
$\Longleftrightarrow a^2-2ax+x^2=a-\sqrt{x}$

after ?

4. Originally Posted by Perelman
$x= a-\sqrt{a-\sqrt{x}}$
$\Longleftrightarrow a-x=\sqrt{a-\sqrt{x}}$
$\Longleftrightarrow a^2-2ax+x^2=a-\sqrt{x}$

after ?
Well if you want to solve for x or a? If it is x then subtract another a from each side and square them again $(a^2-2ax+x^2-a)(a^2-2a+x^2-a) = x$

5. Originally Posted by Henryt999
Well if you want to solve for x or a? If it is x then subtract another a from each side and square them again $(a^2-2ax+x^2-a)(a^2-2a+x^2-a) = x$
Solve the Equation, by following the values of parameter a ( $a \in \mathbb{R^+}$ )

6. Salut !

$x=a-\sqrt{a-\sqrt{x}}$

$\sqrt{x}=\sqrt{a-\sqrt{a-\sqrt{x}}}$

$a-\sqrt{x}=a-\sqrt{a-\sqrt{a-\sqrt{x}}}$

$\sqrt{a-\sqrt{x}}=\sqrt{a-\sqrt{a-\sqrt{a-\sqrt{x}}}}$

$a-\sqrt{a-\sqrt{x}}=a-\sqrt{a-\sqrt{a-\sqrt{a-\sqrt{x}}}}$

$x=a-\sqrt{a-\sqrt{a-\sqrt{a-\sqrt{x}}}}$

and so on ...

Therefore taking the series defined by $u_0 = a$ and $u_{n+1}=f(u_n)$ where $f(x) = a-\sqrt{x}$

x can be seen as the limit (if it exists) of $\left(u_n\right)_n$

If it converges, it is towards l such as $l = f(l)$ which can be solved by a quadratic

At the end $x = \frac{2a+1\pm \sqrt{4a+1}}{2}$ ... if I am not mistaken

It seems to be "minus" but I let you see why !!

All what I have written is to be checked ...