+0

# Unusual simult equations with sqrts

+3
666
4

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 $$abcd$$.

Jan 28, 2021
edited by Melody  May 11, 2021

#3
+9

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} Apr 16, 2021
edited by Divineology  Apr 16, 2021

#1
-4

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

Jan 28, 2021
#2
+2

I'd like to see this one answered.

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

Apr 15, 2021
#3
+9

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
edited by Divineology  Apr 16, 2021
#4
+1

Diveneology:

Thank you very much for your answer.   Apr 16, 2021