Product of roots abcd in 4th degree equation

[juantheron]

If $a,b,c,d$ are distinct 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 $abcd=$

 

[mente oscura]

If $a,b,c,d$ are distinct 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 $abcd=$

Hello.

Spoiler

$$a^4-8a^2-a+11=0$$

$$b^4-8b^2-b+11=0$$

$$c^4-8c^2+c+11=0$$

$$d^4-8d^2+d+11=0$$

Common roots "a" and "b":

$$r_1=-2.3710$$

$$r_2=-1.4551$$

$$r_3=1.2266$$

$$r_4=2.5994$$

Common roots "c" and "d":

$$s_1=2.3710$$

$$s_2=1.4551$$

$$s_3=-1.2266$$

$$s_4=-2.5994$$

There are several solutions product of roots.

A curiosity, jacks. Do you participate in a forum on Spanish?

Regards.

 

[Opalg]

If $a,b,c,d$ are distinct 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 $abcd=$

In order to get a unique solution, I believe the question should say

If $a,b,c,d$ are distinct positive real nos. 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 $abcd=$ ?​

[sp]Then $a,b,c,d$ all satisfy the equation $x = \sqrt{4 \pm\sqrt{5\pm x}}$. Square both sides to get $x^2 - 4 = \pm\sqrt{5\pm x}$. Square both sides again, getting $(x^2-4)^2 = 5 \pm x$, or $x^4 - 8x^2 \pm x + 11 = 0$.

Now it is clear that $x$ is a solution of $x^4 - 8x^2 + x + 11 = 0$ if and only if $-x$ is a solution of $x^4 - 8x^2 - x + 11 = 0$. Also, the sum of the roots of $x^4 - 8x^2 + x + 11$ is $0$, and their product is $11$. Therefore, given that the roots are real, two of them must be positive and two negative. It follows that the numbers $a,b,c,d$ must be the absolute values of the roots of $x^4 - 8x^2 + x + 11$, and therefore their product is $11$.[/sp]