Solving for $abcd$ Given Real Numbers

[anemone]

Let $a, b, c, d$ be real numbers such that \(\displaystyle a=\sqrt{4-\sqrt{5-a}}\), \(\displaystyle b=\sqrt{4+\sqrt{5-b}}\), \(\displaystyle c=\sqrt{4-\sqrt{5+c}}\) and \(\displaystyle d=\sqrt{4+\sqrt{5+d}}\). Calculate $abcd$.

 

[Opalg]

Let $a, b, c, d$ be real numbers such that \(\displaystyle a=\sqrt{4-\sqrt{5-a}}\), \(\displaystyle b=\sqrt{4+\sqrt{5-b}}\), \(\displaystyle c=\sqrt{4-\sqrt{5+c}}\) and \(\displaystyle d=\sqrt{4+\sqrt{5+d}}\). Calculate $abcd$.

[sp]$a$ and $b$ satisfy the equation $(x^2-4)^2 = 5-x$, or $x^4 - 8x^2 + x + 11 = 0\quad(*).$

$c$ and $d$ satisfy the equation $(x^2-4)^2 = 5+x$, or $x^4 - 8x^2 - x + 11 = 0\quad(**).$

But if $x$ satisfies (**) then $-x$ satisfies (*). So the roots of (*) are $a,b,-c,-d.$ Thus $abcd$ is the product of the roots of (*), namely 11.[/sp]

 

[anemone]

[sp]$a$ and $b$ satisfy the equation $(x^2-4)^2 = 5-x$, or $x^4 - 8x^2 + x + 11 = 0\quad(*).$

$c$ and $d$ satisfy the equation $(x^2-4)^2 = 5+x$, or $x^4 - 8x^2 - x + 11 = 0\quad(**).$

But if $x$ satisfies (**) then $-x$ satisfies (*). So the roots of (*) are $a,b,-c,-d.$ Thus $abcd$ is the product of the roots of (*), namely 11.[/sp]

Well done, Opalg! And thanks for participating too! Just so you know, I approached it the same way you did.(Sun)