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Calculate abcd

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anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,756
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

MHB Oldtimer
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Feb 7, 2012
2,725
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 $abcd$.
$a$ 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.
 
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anemone

MHB POTW Director
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Feb 14, 2012
3,756
$a$ 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.
Well done, Opalg! And thanks for participating too! Just so you know, I approached it the same way you did.(Sun)