Solving $(x+2011)(x+2012)(x+2013)(x+2014)+1=0$

[anemone]

Find all real solutions of the equation $(x+2011)(x+2012)(x+2013)(x+2014)+1=0$

 

[Saitama]

Find all real solutions of the equation $(x+2011)(x+2012)(x+2013)(x+2014)+1=0$

Spoiler

Let $x+2013=t$, then the equation is:
$$(t-2)(t-1)t(t+1)+1=0 \Rightarrow t^4-2t^3-2t^2+2t+1=0$$
$$\Rightarrow t^4+t^2+1-2t^3-2t^2+2t=0 \Rightarrow (t^2-t-1)^2=0 \Rightarrow t^2-t-1=0$$
The solutions to the above quadratic are:
$$t=\frac{1+\sqrt{5}}{2},\frac{1-\sqrt{5}}{2}$$
$$\Rightarrow x=\frac{1+\sqrt{5}}{2}-2013,\frac{1-\sqrt{5}}{2}-2013$$

 

[anemone]

Thanks, Pranav for your neat solution!

My solution:

Spoiler

Well, it couldn't be said that it's my method, it was how I remembered how this kind of problem be tackled that I just applied the skill to this problem, so I couldn't really take the credit::)

Let $a=x+\dfrac{2014+2013+2012+2011}{4}=x+2012.5$, then the original equation becomes $(a+1.5)(a+0.5)(a-0.5)(a-1.5)+1=0$ or simply $a^4-2.5a^2+1.5625=0$ and solving it for $a^2$ we get $(a^2-1.25)^2=0$.

Hence, $a=\pm \sqrt{1.25}$ and therefore, $x=\pm \sqrt{1.25}-2012.5$, upon checking, these two solutions do indeed satisfy the original equation and we're done.

 

[kaliprasad]

Thanks, Pranav for your neat solution!

My solution:

Spoiler

Well, it couldn't be said that it's my method, it was how I remembered how this kind of problem be tackled that I just applied the skill to this problem, so I couldn't really take the credit::)

Let $a=x+\dfrac{2014+2013+2012+2011}{4}=x+2012.5$, then the original equation becomes $(a+1.5)(a+0.5)(a-0.5)(a-1.5)+1=0$ or simply $a^4-2.5a^2+1.5625=0$ and solving it for $a^2$ we get $(a^2-1.25)^2=0$.

Hence, $a=\pm \sqrt{1.25}$ and therefore, $x=\pm \sqrt{1.25}-2012.5$, upon checking, these two solutions do indeed satisfy the original equation and we're done.

one can always take credit for applying skill. applying skill is a part of skillset(Clapping)

 

[RLBrown]

Spoiler

Golden Quadratic: $\phi \phi $-$\phi $-1 = 0

1) $\phi $-1 = 1/$\phi $
2) 1 = $\phi \phi $-$\phi $

Given: 1-2+1 = 0
($\phi \phi $-$\phi $)-2+1 = 0 From 2
($\phi $-2)($\phi $+1)+1 = 0 Factoring
($\phi $-2)(1/$\phi $)($\phi $)($\phi $+1)+1 = 0 Multiplicitive Inverse
($\phi $-2)($\phi $-1)($\phi $)($\phi $+1)+1 = 0 From 1

Let x=$\phi $-2013 where $\phi $ is either root of the Golden Quadratic
(x + 2011) (x + 2012) (x + 2013) (x + 2014)+1 = 0 QED

 

[anemone]

Spoiler

Golden Quadratic: $\phi \phi $-$\phi $-1 = 0

1) $\phi $-1 = 1/$\phi $
2) 1 = $\phi \phi $-$\phi $

Given: 1-2+1 = 0
($\phi \phi $-$\phi $)-2+1 = 0 From 2
($\phi $-2)($\phi $+1)+1 = 0 Factoring
($\phi $-2)(1/$\phi $)($\phi $)($\phi $+1)+1 = 0 Multiplicitive Inverse
($\phi $-2)($\phi $-1)($\phi $)($\phi $+1)+1 = 0 From 1

Let x=$\phi $-2013 where $\phi $ is either root of the Golden Quadratic
(x + 2011) (x + 2012) (x + 2013) (x + 2014)+1 = 0 QED

Thanks, RLBrown for participating! You know, I wouldn't have thought to approach it this way and thank you again for showing us the other good method to solve this challenge problem. :)