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# Help

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If $$x=\frac{\sqrt5-3}{2}$$

Find $$x(x-1)(x-2)(x-3)$$

Jun 4, 2018

#1
+974
+2

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
+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
+974
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
+27664
+2

Hmm.

Jun 4, 2018