Solve-the-equation-x-a-a-x-where-a-gt-0-is-a-parameter-

Question Number 149670 by EDWIN88 last updated on 06/Aug/21

S o l v e t h e e q u a t i o n

x = \sqrt{a - \sqrt{a + x}} w h e r e a > 0 i s

a p a r a m e t e r .

Answered by MJS_new last updated on 06/Aug/21

for a, x \in R

{it}^{'} s easy to see that

0 ⩽ x ⩽ a (a - 1)

\Rightarrow a ⩾ 1

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

squaring and transforming 2 times

(beware of false solutions!)

leads to

a^{2} - (2 x^{2} + 1) a + x (x^{3} - 1) = 0

\Rightarrow

a_{1} = x^{2} - x \lor a_{2} = x^{2} + x + 1

testing a_{1} :

x = \sqrt{x^{2} - x - \sqrt{x^{2} - x + x}}

x = \sqrt{x^{2} - 2 x} \Rightarrow x = 0 \Rightarrow a = 0

testing a_{2} :

x = \sqrt{x^{2} + x + 1 - \sqrt{x^{2} + x + 1 + x}}

x = \sqrt{x^{2} + x + 1 - ∣ x + 1 ∣}

x ⩾ 0 \Rightarrow x = \sqrt{x^{2}} always true

\Rightarrow

a = x^{2} + x + 1 \Leftrightarrow x = - \frac{1}{2} \pm \frac{\sqrt{4 a - 3}}{2}

x ⩾ 0 \Rightarrow

★ x = - \frac{1}{2} + \frac{\sqrt{4 a - 3}}{2} ★

Commented by mr W last updated on 07/Aug/21

g r e a t!

Facebook Tweet Pin