Question-183711

Question Number 183711 by Michaelfaraday last updated on 29/Dec/22

Commented by Matica last updated on 29/Dec/22

t o m a t h e m a t i z e

Answered by HeferH last updated on 29/Dec/22

\sqrt[3]{x^{2} \sqrt[3]{x^{2} \sqrt[3]{x^{2} \dots}}} = H

\sqrt[3]{x^{2} H} = H

{H x}^{2} = H^{3}

x = H

\sqrt{x^{2} + x \sqrt{x^{2} + x \sqrt{x^{2} + x \sqrt{\dots}}}} = P

\sqrt{x^{2} + x \cdot P} = P

x^{2} + P x = P^{2}

x^{2} + P x - P^{2} = 0

\frac{x}{P} + 1 - \frac{P}{x} = 0

l e t \frac{x}{P} = t

t + 1 - \frac{1}{t} = 0

t^{2} + t - 1 = 0

t = \frac{- 1 \pm \sqrt{1 - 4 (1) (- 1)}}{2 (1)} = \frac{- 1 \pm \sqrt{5}}{2}

\frac{x}{P} = \frac{- 1 + \sqrt{5}}{2}

\frac{2 x}{\sqrt{5} - 1} = P \Rightarrow

\frac{x}{\frac{2 x}{\sqrt{5} - 1}} = \frac{x}{5} \Rightarrow \frac{2 x}{\sqrt{5} - 1} = 5

x = \frac{5 \sqrt{5} - 5}{2}

Commented by JDamian last updated on 29/Dec/22

I n t h i s s t e p

x^{2} + P x = P^{2}

w h y {d o n}^{'} t y o u s o l v e q u i c k l y a s a q u a d r a t i c

e q u a t i o n ? Y o u h a v e c h a n g e d i n t o a n o t h e r

q u a d r a t i c e q u a t i o n i n t a n d s o l v e d t o g e t P,

w h i c h i s u n n e c e s s a r y .

Commented by Michaelfaraday last updated on 29/Dec/22

t h a n k s s i r

b u t s i r i t h i n k y o u s h o u l d h a v e s o l v e i t

i n f o r m o f q u a d r a t i c e q n f r o m

x^{2} + p x - p^{2} = 0 w h i l e y o u r s o l v i n g f r o m e q n

w h i c h i s n o t n e c e s s a r y .

Commented by HeferH last updated on 29/Dec/22

Y e a h y o u a r e r i g h t, i n m y d e f e n s e

i t w a s v e r y l a t e (00 : 48) w h e n i d i d i t

Commented by Michaelfaraday last updated on 29/Dec/22

o k a y s i r