Let a, b, c, and d be distinct real numbers such that

\(\begin{align*} a &= \sqrt{4 + \sqrt{5 + a}}, \\ b &= \sqrt{4 - \sqrt{5 + b}}, \\ c &= \sqrt{4 + \sqrt{5 - c}}, \\ d &= \sqrt{4 - \sqrt{5 - d}}. \end{align*}\)

Compute \(abcd\).

Divineology Jan 28, 2021

#1**-3 **

The solutions are 2.2192, 1.33938, 1.09044, and 2.4682, and their product is 8.

Guest Jan 28, 2021

#2**+1 **

I'd like to see this one answered.

Divineology, perhaps you can show us how it is done? Or anyone else?

Melody Apr 15, 2021

#3**+8 **

Squaring both sides of the original equation, we get \(a^2 = 4+\sqrt{5+a}.\)

Subtracting 4 from both sides and squaring again gives \((a^2-4)^2 = 5+a.\)

Expanding this out and subtracting 5 + a from both sides, we have \(a^4-8a^2-a+11 = 0.\)

Similar manipulations on the other equations give \(\begin{align*} b^4-8b^2-b+11 &= 0, \\ c^4-8c^2+c+11 &= 0, \\ d^4-8d^2+d+11 &= 0. \end{align*}\)

(Note the signs carefully.)

Let \(f(x)=x^4-8x^2-x+11\). Then a and b are roots of f(x), while c and d are roots of \(x^4-8x^2+x+11\), which is f(-x). It follows that -c and -d are roots of f(x). Since -c and -d are negative and hence distinct from a and b, we have four roots of f(x). Since f(x) has degree four, we know we have found all the roots.

By Vieta's formulas, the constant term of f(x) is the product of the roots. Thus,

\(\begin{align} abcd = (a)(b)(-c)(-d) = \boxed{11}. \end{align}\)

Divineology
Apr 16, 2021