Equation Magic?!

Algebra Level 3

If $x + \dfrac{1}{x} = 2$ , what can be said about the value of $x^n + \dfrac{1}{x^n}$ for any real number $n \neq 1$ ?

The resulting value must be 2 for the rest of

n

's. The values are different for certain

n

's. The signs of the value 2 alternate.

2 solutions

Chew-Seong Cheong
Mar 20, 2017

$\begin{aligned} x + \frac 1x & = 2 & \small \color{#3D99F6} \text{Multiplying both sides by }x \text{ and rearrange.} \\ x^2 - 2x + 1 & = 0 \\ (x-1)^2 & = 0 \\ \implies x & = 1 \end{aligned}$

Since $x=1$ , $\implies x^n + \dfrac 1{x^n} = 1+1 = \boxed{2}$ for all real $n$ .

Md Zuhair
Mar 19, 2017

Actually I thought like,

as $x + \dfrac{1}{x} = 2$

We can rewrite as $\dfrac{x^2+1}{x} = 2$

So , $x^2 + 1 = 2x$

So $(x-1)^2 = 0$

So $x= 1$

Now $x^n=1$ for all n's other than infinity.

So i think the The correct option is the answer .

@Michael Huang is it correct?

Because the solution is unique for $x + \dfrac{1}{x} = 2$ , the possible choice is $x = 1$ , so you are right! :)

Michael Huang - 4 years, 2 months ago

Equation Magic?!

2 solutions

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