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If \(x=\frac{\sqrt5-3}{2}\)

 

Find \(x(x-1)(x-2)(x-3)\)

 Jun 4, 2018
 #1
avatar+981 
+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. 

 Jun 4, 2018
 #2
avatar
+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

 Jun 4, 2018
 #4
avatar+981 
0

Yeah, I have realized. 

 

I was so caught up in writing the solution to the problem, 

 

I just made a careless error. 

 

Thanks for spotting it. 

GYanggg  Jun 4, 2018
 #3
avatar+33603 
+3

Hmm.

 

 Jun 4, 2018

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