<s>[INST] If $a=\sqrt{4+\sqrt{5+a}}\;\;,b=\sqrt{4-\sqrt{5+b}}\;\;,c=\sqrt{4+\sqrt{5-c}}\;\;d=\sqrt{4-\sqrt{5-d}}$. Then $abcd $ If $a,b,c,d$ are $4$ distinct positive real no. such that 
$a=\sqrt{4+\sqrt{5+a}}\;\;,b=\sqrt{4-\sqrt{5+b}}\;\;,c=\sqrt{4+\sqrt{5-c}}\;\;d=\sqrt{4-\sqrt{5-d}}$. 
Then $abcd = $
$\bf{My\; Try}::$ Let here we form an equation $\displaystyle x=\sqrt{4\pm \sqrt{5\pm x}},$ where $x=a,b,c,d$
$\Longrightarrow \displaystyle x^2=4\pm \sqrt{5\pm x}$
$\Longrightarrow \displaystyle \left(x^2-4\right)^2=\left(\sqrt{5\pm x}\right)^2 = (5+x)\;\;,\;\;(5-x)$
$\Longrightarrow \displaystyle x^4+16-8x^2=5+x\;\; ,\;\;  5-x$
$\Longrightarrow \displaystyle x^4-8x^2-x+11=0\;\;,x^4-8x^2+x+11=0$
Now i did not understand how can i solve it.
Help me
Thanks
 [/INST] Hint $a,b,-c,-d$ are the roots of the same equation.
</s>