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Algebra Level 2

If $x + \dfrac{1}{x} = 2$ , then find $x^{99} +\dfrac{1}{x^{99} }$ .

8 2 6 3

2 solutions

Zach Abueg
Feb 3, 2017

$\displaystyle x + \dfrac 1x = 2$

$\displaystyle \dfrac{x^2 + 1}{x} = 2$

$\displaystyle x^2 - 2x + 1 = 0 \Longrightarrow x = 1$

$\displaystyle 1^{99} + \dfrac{1}{1^{99}} = 2$

Hjalmar Orellana Soto
Feb 6, 2017

$x+\frac{1}{x}=2 \Rightarrow x^2+\frac{1}{x^2}=2 \Rightarrow x^n+\frac{1}{x^n}=2$ For all even $n$ $x^{99}+\frac{1}{x^{99}}=(x+\frac{1}{x})(x^{98}+\frac{1}{x^{98}})-(x+\frac{1}{x})=2\cdot 2 -2=2$

Why were you able to generalize $x^n + \frac{1}{x^n} \ \forall$ even $n$ ?

Zach Abueg - 4 years, 3 months ago

In general, under the given condition $x^n+\frac{1}{x^n}=2$ $\forall$ integer $n$ ... but don't know how to prove it.... I'll be thinking about

Hjalmar Orellana Soto - 4 years, 3 months ago

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