Find the Value of $12x^4-2x^3-25x^2+9x+2017$ with $x=\dfrac {\sqrt 5 +1}{4}$

[Albert1]

$x=\dfrac {\sqrt 5 +1}{4}$

please find the value of $12x^4-2x^3-25x^2+9x+2017$

 

[anemone]

My solution:

Spoiler

We're given $x=\dfrac {\sqrt 5 +1}{4}$, and this gives us $x^2=\dfrac {2x +1}{4}\rightarrow4x^2-2x=1$ $\therefore 12x^2-6x=3,\,\,4x+\dfrac{8}{x}=2,\,\,20x^2-10x=5$

We're asked to evaluate $12x^4-2x^3-25x^2+9x+2017$:

First, we let

$12x^4-2x^3-25x^2+9x=k$

Manipulating the equation above algebraically, we see that

$12x^2-2x-25+\dfrac{9}{x}=\dfrac{k}{x^2}$

$(12x^2-6x)+\left(4x+\dfrac{8}{x}\right)+\dfrac{1}{x}-25=\dfrac{k}{x^2}$

$3+2+\dfrac{1}{x}-25=\dfrac{k}{x^2}$

$k=-(20x^2-10x)=-5$

$\therefore 12x^4-2x^3-25x^2+9x+2017=k+2017=-5+2017=2012$

 

[kaliprasad]

Spoiler

we have $4x-1 = \sqrt(5)$

squaring and reordering we get
$16x^2-8x -4-0$
or $4x^2-2x-1 = 0 \cdots (1)$

now deviding $12x^4-2x^3-25x^2+ 9x + 2017$ by $(4x^2-2x-1)$ we find that

$12x^4-2x^3-25x^2+9x+2017=(4x^2-2x-1)(3x^2+x+5) + 2012= 2012$

 

[Greg]

Spoiler

A somewhat tedious but perhaps mildly interesting solution:

\(\displaystyle x=\frac{\sqrt5+1}{4}=\frac12\varphi\) where \(\displaystyle \varphi\) is the golden ratio.

Identity: \(\displaystyle \varphi^2=\varphi+1\)

\(\displaystyle 12x^4-2x^3-25x^2+9x+2017\)

\(\displaystyle =12\left[\frac14(\varphi+1)\right]^2-\frac14(\varphi^2+\varphi)-25\left[\frac14(\varphi+1)\right]+\frac92\varphi+2017\)

\(\displaystyle =\frac34(3\varphi+2)-\frac14(2\varphi+1)-\frac{25}{4}(\varphi+1)+\frac92\varphi+2017\)

\(\displaystyle =\frac94\varphi+\frac32-\frac12\varphi-\frac14-\frac{25}{4}\varphi-\frac{25}{4}+\frac92\varphi+2017\)

\(\displaystyle =2012\)

 

[Albert1]

greg1313 :
yes, very good and very interesting solution !
mine is the same as kaliprasad's solution