Question

In the sequence of non-zero numbers x^1, x^2, x^3, x^4,..., the nth term is x^n. If, from the third term on, each term is the sum of the preceding two terms, what are all possible values of x?

Answer 1

\[x^3=x^2+x\] i guess is what it is saying

Answer 2

and apparently \(x\neq 0\) so you can solve
\[x^3-x^2-x=0\]
\[x(x^2-x-1)=0\] solve
\[x^2-x-1=0\] use quadratic formula

Answer 3

you good with this or need more help?

Answer 4

I'm... still a little bit confused, I think.

Answer 5

ok lets go slow

Answer 6

If, from the third term on, each term is the sum of the preceding two terms

Answer 7

this means the third term, which is \(x^3\) is the sum of the previous two terms, \(x\) and \(x^2\)

Answer 8

so we can write the equation
\[x^3=x^2+x\]

Answer 9

ok, that makes sense now...

Answer 10

wait a second...

Answer 11

ok

Answer 12

i just read back over the first part... quadratic formula...

Answer 13

once you have the equation
\[x^3-x^2-x=0\] you can factor as
\[x(x^2-x-1)=0\]

Answer 14

one solution is \(x=0\) but you are told at the beginning that \(x\) is not zero, so you can now solve
\[x^2-x-1=0\]

Answer 15

right... I think i have an answer, but i'm not entirely sure how to type it... one plus-minus root five, all over two?

Answer 16

Or did I just do something completely different?

Answer 17

yes that is the answer

Answer 18

Ok, thanks! ^_^

Answer 19

it might be worth checking, but is should be right

Answer 20

btw, \(\frac{1+\sqrt{5}}{2}\) is a famous number, sometime denoted as \(\phi\) or the "golden ratio"
if you google it you will see lots

Answer 21

Huh.
I'll remember that. It does look kind of familiar

Answer 22

it is related to fibonacci numbers

Answer 23

If k is real, what is the only real number that could be a multiple root of x^3 +kx+1=0?

Answer 24

oh now i have to think

Answer 25

we can do this the long way, i am not sure there is a simple way

Answer 26

Ok, so now I know the answer I would usually put down in this situation is wrong. So how does the long way work?

Answer 27

expand \((x-a)^2(x-b)\) if the two roots are say \(a\) and \(b\) and then equate

Answer 28

you get
\[(x-a)^2(x-b)=x^3-(2a+b)x^2+(a^2+2ab)x-a^2b\] if my algebra is right

Answer 29

and since there is no \(x^2\) term this tells you \(2a+b=0\)

Answer 30

you also know \(-a^2 b=1\)

Answer 31

Whoa. How did you get that? The part right after you've expanded? I got something totally different. Or maybe I'm just working in circles and getting nowhere... I don't know what i just did. One second...

Answer 32

you have to grind it out and then combine like terms, but i cheated

Answer 33

So, there's nothing in here that says x^3-2ax^2+a^2x-bx^2+2abx-a^2b=0? Or did you get that before combining like terms?

Answer 34

Wait , are there even like terms in here? I think I just blew this way out of proportion...

Answer 35

There's nothing to go with x^3...

Answer 36

yes you are right, on the right track for sure

Answer 37

you got this \[ x^3-2ax^2+a^2x-bx^2+2abx-a^2b\] right?

Answer 38

but what about -2ax^2 and -bx^2? they both have x^2, but would the coefficient be -2ab? and yes, i got that=0

Answer 39

\(-2ax^2-bx^2=(-2a-b)x^2=-(2a+b)x^2\) is the algebra i used to get the coefficient of the \(x^2\) term

Answer 40

oh like solving a literal equation, kind of?

Answer 41

oh, i see what you did, no you don't multiply to get \(-2abx^2\) you add

Answer 42

yeah kind of, but really it is "factoring" as in "factor the \(x^2\) out of the expression
\[-2ax^2-bx^2\]

Answer 43

so i multiplied instead of added, and i didnt even multiply correctly, now that i look at it. oh dear.

Answer 44

similarly for the \(x\) term you get \((a^2+2ab)x\) but that is not really important here

Answer 45

hope it is clear what the real problem is, find the polynomial that has no \(x^2\) term

Answer 46

so, now what about a^2x and -a^2b? that would be... (x-b)a^2, right?

Answer 47

no, you want all in terms of \(x\), not \(a^2\) we just made up the \(a\) as a root

Answer 48

the like terms are in terms of \(x\)

Answer 49

Oh. Lord, Okay. so what would I do with those two terms, then? aren't they like terms?

Answer 50

and \(-a^2b\) is the constants, because there is no \(x\) in it, so we know \(-a^2b=1\)

Answer 51

wait. the like terms are in terms of x. ok. hold on...

Answer 52

k

Answer 53

so do i need to solve x^3-(2a+b)x^2+1=0?

Answer 54

if -a^2b=1?

Answer 55

lets back up a second

Answer 56

good idea.

Answer 57

when you combine like terms (in \(x\)\) you should get
\[(x-a)^2(x-b)=x^3-(2a+b)x^2+(a^2+2ab)x-a^2b\]

Answer 58

and this must be equal to \(x^3+kx+1\)

Answer 59

ok, i see that first part now...

Answer 60

the important point is this: \(x^3+kx+1\) has no \(x^2\) term in it

Answer 61

right.
so... do i set the two expressions equal to each other?

Answer 62

forget about the \(k\) we don't need it
what has to be true is that \(2a+b=0\)

Answer 63

because there is no \(x^2\) term

Answer 64

and also you know the constant is \(-a^2b=1\)

Answer 65

i mean x^3+kx+1=x^3-(2a+b)x^2+(a^2+2ab)x-a^2b?

ok, back up. im lost again.

Answer 66

yes you know
\[ x^3+kx+1=x^3-(2a+b)x^2+(a^2+2ab)x-a^2b\]

Answer 67

so they are equal?

Answer 68

they must be equal, yes, because you are given at the outset that the polynomial is \(x^3+kx+1\)

Answer 69

so now lets concentrate on finding the roots, which are \(a\) and \(b\)

Answer 70

ok. so... how do i do that? unless i just square root it, but that leaves me with a bunch of radicals, right?

Answer 71

now it the time to equate like coefficents

Answer 72

the coefficient of the \(x^2\) term in \(x^3+kx+1\) is \(0\) and therefore
\[2a+b=0\]

Answer 73

ok... so... let me think...

Answer 74

wait, why 0?

Answer 75

\[x^3+kx+1=x^3-(2a+b)x^2+(a^2+2ab)x-a^2b\]

Answer 76

oh x^2. not x^3. got it

Answer 77

we also see that \(-a^2b=1\)

Answer 78

how do we see that?

Answer 79

look above

Answer 80

oh. OH. i think im understanding a little better.

Answer 81

if you plugged in 0 for x and solved, that'd be it, right? 1=-a^2b?

Answer 82

\[x^3+kx+1=x^3-\overbrace{(2a+b)}^0x^2+\overbrace{(a^2+2ab)}^kx\overbrace{-a^2b}^1\]

Answer 83

yes, that is another way to see it!

Answer 84

aha! makes sense now.

Answer 85

so now we can solve
\[2a+b=0\]
\[-a^2b=1\] for \(a\) and \(b\)

Answer 86

so now my equation looks like x^3+kx+1=x^3+(a^2+2ab)x+1, right? and i solve that?

Answer 87

we need the roots, that is, we need \(a\) and \(b\)

Answer 88

ok, so im still not sure how to do that without ending up with a bunch of radicals in the equation. or can i just get rid of those easily enough after i find the roots? no, bc then i'd be going in a circle...

Answer 89

>.<

Answer 90

now it is just algebra, if \(2a+b=0\) then \(b=-2a\) and if \(-a^2b=1\) if you replace \(b\) by \(-2a\) you get
\[-a^2(-2a)=1\\
2a^3=1\\
a=\frac{1}{\sqrt[3]{2}}\]

Answer 91

at least that is what i got,

Answer 92

oh. just algebra. my bad; i over-thought again.
ok, so... once you get there, you'd multiply by the denominator, right?

Answer 93

lets go back up and read the question

Answer 94

took a while
it says this
"If k is real, what is the only real number that could be a multiple root of x^3 +kx+1=0"

Answer 95

ugh. what have i done? i thought that's what you did to get a radical out of the denominator...?

Answer 96

oh if you want, sure go ahead
i just wanted to make sure we knew we had the answer

Answer 97

what?!

Answer 98

i think we found it, though, it should be
\[x=\frac{1}{\sqrt[3]{2}}\]