Solve Equation: $1-2a+2a^2+2a^3+a^4$

[anemone]

Solve the equation $1-2a+2a^2+2a^3+a^4=(a^3+a)\sqrt{\dfrac{1-a^2}{a}}$.

 

[Albert1]

Solve the equation $1-2a+2a^2+2a^3+a^4=(a^3+a)\sqrt{\dfrac{1-a^2}{a}}$.

Spoiler

$a= -1\pm \sqrt 2$

 

[Albert1]

Solve the equation $1-2a+2a^2+2a^3+a^4=(a^3+a)\sqrt{\dfrac{1-a^2}{a}}---(1)$.

Spoiler

recombination and simplification of(1) we get:
$(a^2+1)^2-2a(1-a^2)=(a^2+1)\sqrt{a(1-a^2)}$
let $a^2+1=x,\sqrt{a(1-a^2)}=y$
and we have $x^2-2y^2=xy$
or
$(x+y)(x-2y)=0---(2)$
from (2) we get :$x=2y$
$a^2+1=2\sqrt{a(1-a^2}--(3)$
square both side of (3) we have :
$a^4+4a^3+2a^2-4a+1=0$
or
$(a^2+2a)^2-2(a^2+2a)+1=0$
$\therefore a^2+2a=1$
and $a=-1\pm\sqrt 2$

 

[anemone]

Spoiler

recombination and simplification of(1) we get:
$(a^2+1)^2-2a(1-a^2)=(a^2+1)\sqrt{a(1-a^2)}$
let $a^2+1=x,\sqrt{a(1-a^2)}=y$
and we have $x^2-2y^2=xy$
or
$(x+y)(x-2y)=0---(2)$
from (2) we get :$x=2y$
$a^2+1=2\sqrt{a(1-a^2}--(3)$
square both side of (3) we have :
$a^4+4a^3+2a^2-4a+1=0$
or
$(a^2+2a)^2-2(a^2+2a)+1=0$
$\therefore a^2+2a=1$
and $a=-1\pm\sqrt 2$

Thanks for participating, Albert...

Spoiler

but this problem only has one unique solution...(Smile)

 

[Albert1]

$a=-1+\sqrt 2$

$"a=-1-\sqrt 2"$ does not fit , should be deleted