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Algebra Level 2

If $x^2-3x+1=0$ , then what is the value of $x^5+x^{-5}$ ?

119

123

125

122

2 solutions

Chew-Seong Cheong
Apr 9, 2021

Given that

$\begin{aligned} x^2 - 3x + 1 & = 0 & \small \blue{\text{Divide both sides by }x} \\ x - 3 + \frac 1x & = 0 \\ \implies x + \frac 1x & = 3 \\ \left(x + \frac 1x \right)^3 & = 3^3 \\ x^3 + \blue{3x + \frac 3x} + \frac 1{x^3} & = 27 \\ x^3 + \blue{3\cdot 3} + \frac 1{x^3} & = 27 \\ \implies x^3 + \frac 1{x^3} & = 18 \\ \left(x + \frac 1x \right)^5 & = 3^5 \\ x^5 + 5x^3 + 10x + \frac {10}x + \frac 5{x^3} + \frac 1{x^5} & = 243 \\ \implies x^5 + \frac 1{x^5} & = 243 - 5 \left(x^3 + \frac 1{x^3} \right) - 10 \left(x + \frac 1x \right) \\ & = 243 - 5 \cdot 18 - 10 \cdot 3 = \boxed {123} \end{aligned}$

Nice solution! But it should be $\dfrac{1}{x^3}$ in lines 5, 6 and 7

Aditya Mittal - 2 months ago

Thanks. I copied and pasted without checking.

Chew-Seong Cheong - 2 months ago

Chris Lewis
Apr 9, 2021

This can be solved using binomial expansions of $\left(x+x^{-1}\right)^n$ , but here's an alternative idea.

First note that if $x$ is one root of the given equation, the other is $x^{-1}$ . Also, dividing the equation through by $x$ (we can, it's clearly non-zero) and rearranging gives $x+x^{-1}=3$

Now, let $a_n=x^n+x^{-n}$ , and note that this is the form of a solution to a recurrence relation; in fact, it's a solution to $a_{n+2}-3a_{n+1}+a_n=0$

We have $a_0=1+1=2$ and $a_1=3$ (from above); so we can just use the recurrence to calculate

$n$	$a_n$
$0$	$2$
$1$	$3$
$2$	$7$
$3$	$18$
$4$	$47$
$5$	$123$

so the answer we need is $\boxed{123}$ .

Bonus fact: that if $F_n$ is the $n^\text{th}$ Fibonacci number (starting with $F_0=F_1=1$ ), then $a_n=F_{2n-2}+F_{2n}$ for $n>0$ ; for example, $a_5=34+89=F_8+F_{10}$ . Why does this work?

Chris Lewis - 2 months ago

I'm not sure I understand.

Aditya Mittal - 1 month, 4 weeks ago

Which part?

Chris Lewis - 1 month, 4 weeks ago

$a_n = x^n + x^{-n}$ and everything after that.

Aditya Mittal - 1 month, 4 weeks ago

@Aditya Mittal – Ah, OK. A recurrence relation is a way of defining a sequence; for example you might be given $a_0=2$ and the relation $a_{n+1}=3a_{n}$ , and you could work out that the sequence is $2,6,18,54,\cdots$ . Not only that, but you can see that a general formula for this sequence is $a_n=2\cdot 3^n$ .

Another recurrence relation you've probably already seen is the Fibonacci sequence, given by $F_1=F_0=1$ and $F_{n+2}=F_{n+1}+F_n$ .

With recurrence relations of this type, where the only terms are multiples of terms of the sequence, the solution comes from substituting $a_n=C t^n$ where $C$ and $t$ are numbers to be found. As an example, let's try $a_0=0$ , $a_1=1$ and $a_{n+2}=5a_{n+1}-6a_n$ .

The first few terms are $0,1,5,19,65$ . Let's try the substitution: $Ct^{n+2}=5Ct^{n+1}+Ct^n$

Now we can divide through by $Ct^n$ to get $t^2=5t+6$

which is just a quadratic with solutions $t=2$ and $t=3$ . To combine solutions, we just add them up to get $a_n=C_1 2^n + C_2 3^n$

To find the $C_1,C_2$ values, use the given terms: $a_0=0=C_1+C_2$ and $a_1=1=2C_1+3C_2$

Solving these we get $C_1=-1$ , $C_2=1$ so the solution is $a_n=3^n-2^n$ (and you can check this works).

If you look up recurrence relations, you'll find loads of examples like this. When you've seen a few, you'll recognise the pattern that shows why they're useful in this problem.

Hope that helps!

Chris Lewis - 1 month, 4 weeks ago

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