#1**+2 **

Hey Rollingblade!

We have two equations we can manipulate here:

First, let's change the second one:

\(x(x-1)(x-2)(x-3)\\ =[x(x-3)][(x-1)(x-2)]\\ =(x^2-3x)(x^2-3x+3)\)

Now, we can change the first one:

\(x=\frac{\sqrt5-3}{2}\\ 2x+3=\sqrt5\\ (2x+3)^2=5\\ 4x^2+12x+4=0\\ x^2+3x=-1\)

We now found the value of \(x^2+3x\), and can plug this into the original equation.

\((x^2-3x)(x^2-3x+3)=-1(-1+3)=\boxed{-2}\)

I hope this helped,

Gavin.

GYanggg
Jun 4, 2018

#2**+1 **

well that was a long answer, but sorry gavin, x^2-3x does not equal x^2+3x

you could just sub in [(sqrt5-3)/2] everywhere you see x then simplify

whether you change the equation like gave did, may or may not help

[(sqrt5-3)/2] * [(sqrt5-3)/2 - 2/2] * [(sqrt5-3)/2 - 4/2] * [(sqrt5-3)/2 - 6/2]

{[(sqrt5-3)/2] * [(sqrt5-5)/2]} * {[(sqrt5-7)/2] * [(sqrt5-9)/2]}

{[(-8sqrt5-10)/4]} * {[(-16sqrt5-58)/4]}

[(624sqrt5+1220)/16]

or 99.17 rounded

Guest Jun 4, 2018