Simultaneous Equations Challenge II

[anemone]

Solve for real solutions of the system of equations below:

$a(\sqrt{b}+b)=\sqrt{1-a}(\sqrt{a}+\sqrt{1-a})$

$32a(a^2-1)(2a^2-1)^2+b=0$

 

[anemone]

Hint:

Spoiler

Try to think from the perspective of injective function and also use the trigonometric substitution for the second equation.(Nod)

 

[anemone]

Solution of other:

Spoiler

First equation implies $a\in (0,\,1]$ and $b\ge 0$ and may then be written as $b+\sqrt{b}=\dfrac{1-a}{a}+\sqrt{\dfrac{1-a}{a}}$.

Since $a+\sqrt{a}$ is injective, this implies $b=\dfrac{1-a}{a}$ and the second equation becomes $a\ne 0$ and $32a^2(a^2-1)(2a^2-1)^2=a-1$.

Setting $a=\cos t$, this is $\cos 8t=\cos t$ and since $a>0$, $a\in \left( \cos \dfrac{4\pi}{9},\,\cos \dfrac{2\pi}{9},\,\cos \dfrac{2\pi}{7},\,1\right)$.

Hence the 4 solutions are shown below:

$(a,\,b)=(\cos \dfrac{4\pi}{9},\,\dfrac{1}{\cos \dfrac{4\pi}{9}}-1),\,(\cos \dfrac{2\pi}{9},\,\dfrac{1}{\cos \dfrac{2\pi}{9}}-1),\,(\cos \dfrac{2\pi}{7},\,\dfrac{1}{\cos \dfrac{2\pi}{7}}-1),\,(1,\,0)$.