MHB Find Real Root: $x^5-10x^3+20x-12=0$

[anemone]

Find a real root of $x^5-10x^3+20x-12=0$.

 

[Albert1]

Find a real root of $x^5-10x^3+20x-12=0$.

Spoiler

let: $f(x)=x^5-10x^3+20x-12=0---(1)$
then
$f'(x)=5x^4-30x^2+20$
we know if $x\geq 3$ then (1) no solution
$f(2.83)<0,f(3)>0$
there is a soluion $2.83<x<3$
using Newton's Method:
$x=r - \dfrac{f(r)}{f'(r)}$
if $r=2.83$
we get one solution $x\approx 2.835263$

 

[anemone]

Spoiler

let: $f(x)=x^5-10x^3+20x-12=0---(1)$
then
$f'(x)=5x^4-30x^2+20$
we know if $x\geq 3$ then (1) no solution
$f(2.83)<0,f(3)>0$
there is a soluion $2.83<x<3$
using Newton's Method:
$x=r - \dfrac{f(r)}{f'(r)}$
if $r=2.83$
we get one solution $x\approx 2.835263$

Thanks for participating, Albert! Yes, that is right! Well done!

But what if we want to find the exact root?(Happy)

 

[MarkFL]

Thanks for participating, Albert! Yes, that is right! Well done!

But what if we want to find the exact root?(Happy)

I considered posting the result from Newton's method, but something told me you want the exact value...:D

 

[anemone]

I considered posting the result from Newton's method, but something told me you want the exact value...:D

Aww...I must apologize for not being very clear in the problem ...sorry!(Tmi)

 

[Greg]

Spoiler

It's a DeMoivre quintic. The real root is $$8^{1/5}+\frac{2}{8^{1/5}}$$

 

[Albert1]

Thanks for participating, Albert! Yes, that is right! Well done!

But what if we want to find the exact root?(Happy)

we can never get the exact value of :
$\pi,e,\sqrt 2,\sqrt [5] 8---$ right ?

 

[MarkFL]

we can never get the exact value of :
$\pi,e,\sqrt 2,\sqrt [5] 8---$ right ?

It is true that we can never write the exact value of irrational numbers in decimal form, but giving their symbol or radical notation represents their exact values. :D

 

[Albert1]

Find a real root of $x^5-10x^3+20x-12=0---(1)$.

there is only one real solution k for (1)
if all the solutions of (1) are:
$a\pm bi$,$c\pm di$ and k(real) then :
linear equations:
$2a+2c+k=0-----(1)$
-----
-----
-----
$(a^2+b^2)(c^2+d^2)k=12---(5)$
we will have 5 equations and 5 valuables .it is too complicated ,and should be solved by computer

 

[Greg]

Actually, general solutions to several types of quintic equations were found decades before computers were invented.

 

[anemone]

Thank you MarkFL, greg1313 and Albert for all of the responses...

Solution of other:

Spoiler

Use the substitution where $x=2\sqrt{2}\cosh y$, we see that

$x^5-10x^3+20x-12=0$ becomes

$(2\sqrt{2}\cosh y)^5-10(2\sqrt{2}\cosh y)^3+20(2\sqrt{2}\cosh y)-12=0$

$128\sqrt{2}(\cosh y)^5-160\sqrt{2}(\cosh y)^3+40\sqrt{2}(\cosh y)-12=0$

$\dfrac{128\sqrt{2}(\cosh y)^5}{8\sqrt{2}}-\dfrac{160\sqrt{2}(\cosh y)^3}{8\sqrt{2}}+\dfrac{40\sqrt{2}(\cosh y)}{8\sqrt{2}}=\dfrac{12}{8\sqrt{2}}$

$16\cosh^5 y-20\cosh^3 y+5\cosh y=\dfrac{3}{2\sqrt{2}}$

Using the multiple angle formula for hyperbolic $\cosh$ function we get

$(10\cosh y+5\cosh 3y+\cosh 5y)-(15\cosh y+5\cosh 3y)+5\cosh y=\dfrac{3}{2\sqrt{2}}$

$5\cosh 3y=\dfrac{3}{2\sqrt{2}}$

According to the definition $\cosh^{-1} a=\ln (a\pm\sqrt{a^2-1})$, we find $5y=\ln \left(\dfrac{3\sqrt{2}}{4}\pm\dfrac{\sqrt{2}}{4}\right)=\pm\dfrac{\ln 2}{2}\implies y=\pm 0.1\ln 2$

Again, by the definition $\cosh a=\dfrac{e^{a}+e^{-a}}{2}$, we have

$\begin{align*}x&=2\sqrt{2}\cosh y\\&=2\sqrt{2}\left(\dfrac{e^{0.1\ln 2}+e^{-0.1\ln 2}}{2}\right)\\&=\sqrt{2}(2^{0.1}+2^{-0.1})\\&=2^{0.6}+2^{0.4}\end{align*}$

 

[Opalg]

Find a real root of $x^5-10x^3+20x-12=0$.

This is just to explain how the hint given by greg1313 in comment #6 leads to the solution given by anemone.
[sp]
We look for a solution of the form $x = t + \dfrac at$. Then

$x^5 = t^5 + 5at^3 + 10 a^2t + 10a^3t^{-1} + 5a^4t^{-3} + a^5t^{-5}.$

Also, $x^3 = t^3 + 3at + 3a^2t^{-1} + a^3t^{-3}.$

Therefore $x^5 - 5ax^3 = t^5 - 5a^2t - 5a^3t^{-1} + a^5t^{-5},$ and $x^5 - 5ax^3 + 5a^2x = t^5 + a^5t^{-5}.$

Comparing that with the given equation $x^5-10x^3+20x-12=0$, you see that if we take $a=2$ then we get $x^5-10x^3+20x = t^5 + 32t^{-5}.$

So we want $t^5 + 32t^{-5} = 12$, or $t^{10} - 12 t^5 + 32 = 0.$ That is a quadratic in $t^5$, with solutions $t^5 = 4$ or $8$. Thus $t = 2^{2/5}$ or $2^{3/5}$. Substitute either of those values into the equation $x = t + \dfrac 2t$ and you get the solution $x= 2^{2/5} + 2^{3/5}$ (which is also the solution stated by greg1313).[/sp]