Consider an equation #log_2 (alpha^2-16alpha^3 +66) + sqrt(4beta^4 -8beta^2 +13)+ | (gamma/3-2)| = 4#, Find the numbers of ordered triplets #(alpha, beta, gamma)#? Also find the sum of all possible values of the product #alpha beta gamma#?

The question should have had #alpha^6# instead of #alpha^2# (checked against original question sheet).

Given:

#log_2(alpha^6-16alpha^3+66)+sqrt(4beta^4-8beta^2+13)+abs(gamma/3-2) = 4#

Let us look at each subexpression in turn:

#color(white)()#
(#bb alpha#):

#alpha^6-16alpha^3+66 = (alpha^3)^2-16(alpha^3)+64+2#

#color(white)(alpha^6-16alpha^3+66) = (alpha^3-8)^2+2#

#color(white)(alpha^6-16alpha^3+66) >= 2#

taking the minimum value #2# only when #alpha^3=8#, that is when #alpha=2#.

So:

#log_2(alpha^6-16alpha^3+66) >= log_2 2 = 1#

only taking the minimum value #1# when #alpha=2#.

#color(white)()#
(#bb beta#):

#4beta^2-8beta^2+13 = 4beta^2-8beta^2+4+9#

#color(white)(4beta^2-8beta^2+13) = 4((beta^2)^2-2beta^2+1)+9#

#color(white)(4beta^2-8beta^2+13) = 4(beta^2-1)^2+9#

taking the minimum value #9# when #beta^2 = 1#, i.e. when #beta = +-1#.

Hence:

#sqrt(4beta^2-8beta^2+13) >= sqrt(9) = 3#

taking the minimum value #3# when #beta = +-1#.

#color(white)()#
(#bb gamma#):

#abs(gamma/3-2)#

takes its minimum possible value #0# when:

#gamma/3-2 = 0#

That is, when #gamma = 6#

#color(white)()#
Sum:

So the minimum possible value of:

#log_2(alpha^6-16alpha^3+66)+sqrt(4beta^4-8beta^2+13)+abs(gamma/3-2)#

is #1+3+0 = 4#, only occuring when #alpha=2#, #beta=+-1# and #gamma=6#.

So the only possible solutions of the original equation are:

#(alpha, beta, gamma) = (2, +-1, 6)#

Hence, the sum of all possible values of #alphabetagamma# is:

#(2*1*6)+(2*(-1)*6) = 12-12=0#

We have a relationship as

#f(alpha)+g(beta)+p(gamma) = 4#

with

%%%%%%%%%%%%%%%%%%%
#f(alpha)=log_2(alpha^2 - 16 alpha^3 + 66)#
#g(beta) = sqrt(4 beta^4 - 8 beta^2 + 13)#
#p(gamma) = abs(gamma/3-2)#
%%%%%%%%%%%%%%%%%%%

Considering

#f(alpha)=log_2(alpha^2 - 16 alpha^3 + 66)#

the conditions on #f(alpha)# are:

#alpha^2 - 16 alpha^3 + 66 > 0#
#0 le f(alpha) le 4#

or

#1.48314 le alpha < 1.62487#

The conditions for #g(beta)# are:

#4 b^4 - 8 b^2 + 13 ge 0#
#0 le g(beta) le 4#

for

#-sqrt[1/2 (2 + sqrt[7])] le beta le sqrt[1/2 (2 + sqrt[7])]#

The conditions for #p(gamma)# are:

#0 le p(gamma) le 4#

giving

#-6 le gamma le 18#

Attached a plot showing the variety

#f(alpha)+g(beta)+p(gamma) = 4#