Solve-for-real-numbers-5-5-x-1-5-5-5-x-1-5-2-1-5-

Question Number 154804 by mathdanisur last updated on 21/Sep/21

Solve for real numbers :

\sqrt[5]{5 \sqrt{5} + x} - \sqrt[5]{5 \sqrt{5} - x} = \sqrt[5]{2}

Commented by 7770 last updated on 21/Sep/21

x = 11

Answered by TheHoneyCat last updated on 21/Sep/21

let f : {\begin{cases} R & \to & R \\ x & {(5 \sqrt{5} + x)}^{1 / 5} \end{cases}

Re - writting the equation we get :

\sqrt[5]{2} = {(5 \sqrt{5} + x)}^{1 / 5} - {(5 \sqrt{5} - x)}^{1 / 5}

= f (x) - f (- x)

very obviously f \in D^{1} (R) as a composition of C^{\infty} (R) ones

a c t u a l l y, t o b e r i g o r o u s R s o u l d b e r e p l a c e d b y R - {5 \sqrt{5}} c a u s e \sqrt[n]{} i s n e v e r d e r i v a b l e i n 0 \dots b u t y o u g e t t h e p o i n t

\frac{d f}{d x} = \frac{d (5 \sqrt{5} + x)}{d x} \times \frac{d (x^{1 / 5})}{d x} \circ (x 5 \sqrt{5} + x)

= x \frac{1}{5} {(5 \sqrt{5} + x)}^{- 4 / 5}

= x \frac{1}{5} {({(5 \sqrt{5} + x)}^{2})}^{- 2 / 5}

\forall x \in R - {- 5 \sqrt{5}} {(5 \sqrt{5} + x)}^{2} > 0

so \forall x \in R - {- 5 \sqrt{5}} {({(5 \sqrt{5} + x)}^{2})}^{- 2 / 5} > 0

hence \frac{d f}{d x} > 0 (o v e r R - {5 \sqrt{5}}

of course this gives \frac{d f}{d x} \circ (x - x) < 0 (o v e r R - {- 5 \sqrt{5}})

since f and (therefor x f (x) - f (- x)) is strictly increasing

it as 1 solution or less

considering x = 11

{({(5^{3 / 2} + 11)}^{1 / 5} - {(5^{3 / 2} - 11)}^{1 / 5})}^{5}

Prime causes double exponent: use braces to clarify

i t s a p a i n, s o I {w o n}^{'} t s h o w i t h e r e . S o r r y

Hence x = 11_{}

Commented by TheHoneyCat last updated on 21/Sep/21

s i n c e t h e c a l c u l a t i o n w a s “ a {p a i n}^{″} y o u m i g h t w o n d e r h o w I f o u n d t h a t x = 11 w o u l d w o r k i n t h e f i r s t p l a c e

d i d I s t u b b l e u p o n i t b y p u r e l u c k ?

w e l l, n o t a t a l l, a c t u a l y t h i s w h o l e p r o o f i s t h e o p p o s i t e o f w h a t I d i d b y m y s e l f

I a c t u a l y p l o t t e d t h e c u r v e o n a g r a p h a n d f o u n d (t o a v e r y g o o d p r e c i s i o n) t h a t t h e o n l y s o l u t i o n w a s i n d e e d 11

I t h e n j u s t h a d t o p r o o v e t h e u n i c i t y a n d v e r i f y t h a t 11 w a s n o t j u s t a n a p p r o x i m a t i o n

s e e t h e f o l o w i n g p i c t u r e s I p l o t t e d .

Commented by TheHoneyCat last updated on 21/Sep/21

Here is the equation (t w i s t e d t o g e t a n “ = 0^{″} s t a t e m e n t)

Commented by TheHoneyCat last updated on 21/Sep/21

Commented by TheHoneyCat last updated on 21/Sep/21

and here is the derivative

(I w o n t e d t o m a k e s u r e i t w a s i n d e e d > 0

i f i t {h a d n}^{'} t b e e n, I w o u l d h a v e n o t t r i e d t r o u g h t t h i s m e t h o d)

Commented by TheHoneyCat last updated on 21/Sep/21

Commented by TheHoneyCat last updated on 21/Sep/21

Now, {here}^{'} s a real hard question,

both geogebra and maple, when asked

to solve this, give an exact value (11)

But neither one of them does tel me how

they found that it was an exact value

Does this means they are just too lazy to

tel me, but there DOES exist a way to find

it without first knowing it (i e w i t h o u t a

c o m p u t e r) ? Or are they just pretending

that the value is exact because it is coded

in such a way that 11.000000000000000001

is showed as 11 (i e t h e v a l u e s h o w n i s n o t

f o r m a l a y c a l c u l a t e d, a n d i t s j u s t a f l o a t)

This might be a very “ {m e t h a}^{″} question

for it is at the fronteer of math, computer science

and raises the (a l m o s t p h i l o s o p h i c a l)

question : what does it mean for an equation

to be solvable ?

Please if you have the answere, do give it,

this seems super interesting

: -)

Commented by mathdanisur last updated on 21/Sep/21

let a = 5 \sqrt{5} + x; b = 5 \sqrt{5} - x

then a - b = \sqrt[5]{2}; a^{5} - b^{5} = 2 x and a^{5} + b^{5} = 10 \sqrt{5}

then \frac{a}{\sqrt[5]{2}} - \frac{b}{\sqrt[5]{2}} = 1; {(\frac{a}{\sqrt[5]{2}})}^{5} - {(\frac{b}{\sqrt[5]{2}})}^{5} = x and {(\frac{a}{\sqrt[5]{2}})}^{5} + {(\frac{b}{\sqrt[5]{2}})}^{5} = 5 \sqrt{5}

let \frac{a}{\sqrt[5]{2}} = p and \frac{b}{\sqrt[5]{2}} = q

we have p - q = 1 (1); p^{5} - q^{5} = x (2); p^{5} + q^{5} = 5 \sqrt{5} (3)

also we knov p^{5} + q^{5} = (p + q) [{(p - q)}^{2} ({(p - q)}^{2} + 4 pq) + {2 p}^{2} q^{2} - pq ({(p - q)}^{2} + pq)]

now 5 \sqrt{5} = (p + q) (p^{2} q^{2} + 3 pq + 1)

squaring 125 = {(p + q)}^{2} {(p^{2} q^{2} + 3 pq + 1)}^{2}

5^{3} = (1 + 4 pq) {(p^{2} q^{2} + 3 pq + 1)}^{2} \Leftrightarrow pq = 1

doing {(3)}^{2} - {(2)}^{2}

\Rightarrow 4 {(pq)}^{5} = 125 - x^{2} \Leftrightarrow x^{2} = 121 \Leftrightarrow x = \pm 11

since x = - 11 does not satisfy

we get x = 11

Commented by mathdanisur last updated on 21/Sep/21

Dear S er, thanks for the amazing and

interesting questions

Commented by MJS_new last updated on 22/Sep/21

{it}^{'} s not a proper proof

in line 9 you get p q = 1

but p = \frac{a}{\sqrt[5]{2}} = \frac{11 + 5 \sqrt{5}}{\sqrt[5]{2}} \land q = \frac{b}{\sqrt[5]{2}} = \frac{- 11 + 5 \sqrt{5}}{\sqrt[5]{2}}

\Rightarrow p q = 2^{8 / 5} \neq 1

Answered by MJS_new last updated on 22/Sep/21

how to omit the \sqrt[5]{\dots}

[(*) marks where we introduce false solutions]

a^{\frac{1}{5}} + b^{\frac{1}{5}} = c^{\frac{1}{5}}

a + 5 a^{\frac{4}{5}} b^{\frac{1}{5}} + 10 a^{\frac{3}{5}} b^{\frac{2}{5}} + 10 a^{\frac{2}{5}} b^{\frac{3}{5}} + 5 a^{\frac{1}{5}} b^{\frac{4}{5}} + b = c (*)

5 a^{\frac{1}{5}} b^{\frac{1}{5}} (a^{\frac{3}{5}} + 2 a^{\frac{2}{5}} b^{\frac{1}{5}} + 2 a^{\frac{1}{5}} b^{\frac{2}{5}} + b^{\frac{3}{5}}) = c - a - b

a^{\frac{3}{5}} + 2 a^{\frac{2}{5}} b^{\frac{1}{5}} + 2 a^{\frac{1}{5}} b^{\frac{2}{5}} + b^{\frac{3}{5}} = \frac{c - a - b}{5 a^{\frac{1}{5}} b^{\frac{1}{5}}}

\underset{= c^{\frac{1}{5}}}{\underset{⏟}{(a^{\frac{1}{5}} + b^{\frac{1}{5}})}} (a^{\frac{2}{5}} + a^{\frac{1}{5}} b^{\frac{1}{5}} + b^{\frac{2}{5}}) = \frac{c - a - b}{5 a^{\frac{1}{5}} b^{\frac{1}{5}}}

a^{\frac{2}{5}} + a^{\frac{1}{5}} b^{\frac{1}{5}} + b^{\frac{2}{5}} = \frac{c - a - b}{5 a^{\frac{1}{5}} b^{\frac{1}{5}} c^{\frac{1}{5}}}

now this is what I found

5 {(x^{2} + x y + y^{2})}^{5} = x^{10} + y^{10} + {(x + y)}^{10} + 3 (x^{5} {(x + y)}^{5} + y^{5} {(x + y)}^{5} - x^{5} y^{5})

in our case x = a^{\frac{1}{5}} \land y = b^{\frac{1}{5}} \land (x + y) = c^{\frac{1}{5}}

\Rightarrow

5 {(a^{\frac{2}{5}} + a^{\frac{1}{5}} b^{\frac{1}{5}} + b^{\frac{2}{5}})}^{5} = a^{2} + b^{2} + c^{2} + 3 (a c + b c - a b)

so we can go on

{(a^{\frac{2}{5}} + a^{\frac{1}{5}} b^{\frac{1}{5}} + b^{\frac{2}{5}} = \frac{c - a - b}{5 a^{\frac{1}{5}} b^{\frac{1}{5}} c^{\frac{1}{5}}})}^{5} (*)

\frac{1}{5} (a^{2} + b^{2} + c^{2} + 3 (a c + b c - a b)) = \frac{{(c - a - b)}^{5}}{3125 a b c}

now a = x + 5 \sqrt{5} \land b = x - 5 \sqrt{5} \land c = 2

- \frac{x^{2} - 12 x - 629}{5} = \frac{- 16 {(x - 1)}^{5}}{3125 (x^{2} - 125)}

x^{5} - \frac{705}{16} x^{4} + \frac{1915}{4} x^{3} + \frac{235545}{8} x^{2} - \frac{234355}{4} x - \frac{49140641}{16} = 0

49140641 = 11^{2} \times 101 \times 4021

and luckily x = 11 is a solution

no other solution \in R

Commented by mathdanisur last updated on 22/Sep/21

Creativ solution S er, thank you

for your attention