Algebra #13

by sjaelee, Aug 16, 2011, 9:38 AM

Quote:
Find $abcd$ if \begin{align*}a &=\sqrt{4+\sqrt{5+a}},\\ b &=\sqrt{4-\sqrt{5+b}},\\ c &=\sqrt{4+\sqrt{5-c}},\\ d &=\sqrt{4-\sqrt{5-d}}\end{align*}

We start by finding polynomials to solve for the variables?:

$a=\sqrt{4+\sqrt{5+a}}$, squaring gives

$a^2=4+\sqrt{5+a}$, subtracting four and squaring gives

$a^4-8a^2+16=5+a$, so $a^4-8a^2-a+11=0$. No easy roots, but we can make

similiar equations and process with the other variables:

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

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

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

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

Oh No! Wait! We know $a,b$ are roots of the same polynomial, but $c,d$ differ in

the single degree term. Now we note plugging in $-c$ into the polynomial for $a$

gives the polynomial for $c$, where $c$ is a root! Same with $d$! So the roots of

the polynomial for a are:$(a,b,-c,-d)$.

By Vieta's, $11=\frac{(a)(b)(-c)(-d)}{1}$, so

$abcd=11$
This post has been edited 3 times. Last edited by djmathman, Apr 6, 2015, 2:48 AM
Reason: align tags in new site

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