Author |
Topic: Two real numbers (Read 2968 times) |
|
NickH
Senior Riddler
Gender: 
Posts: 341
|
 |
Two real numbers
« on: Oct 3rd, 2002, 2:08pm » |
 Quote  Modify
|
The sum of the reciprocals of two real numbers is -1, and the sum of their cubes is 4. What are the numbers?
Nick
|
| « Last Edit: Sep 20th, 2003, 6:47pm by Icarus » |
 IP Logged |
Nick's Mathematical Puzzles
|
|
|
Brett Danaher
Guest
|
|
Re: NEW PUZZLE: Two real numbers
« Reply #1 on: Oct 4th, 2002, 7:08am » |
Quote Modify
 Remove
|
I've got an approximate answer
-.61803 and 1.61803
Now, I have a funny feeling that there might be a more exact answer out there involving roots. So if anyone wants to find it, just solve this equation for me.
x^6+3x^5+3x^4-4x^3-12x^2-12x-4=0
I can't remember how to solve n-degree polynomials without resorting to a solver, which gives decimal approximates. Anyone remember the method? I think you might find that polynomial difficult to factor.  Anyway, the solution to that polynomial will yield one of the real numbers - simply use the original conditions to find the other.
|
|
IP Logged |
|
|
|
James Fingas
Uberpuzzler
Gender:
Posts: 949
|
|
Re: NEW PUZZLE: Two real numbers
« Reply #2 on: Oct 4th, 2002, 7:16am » |
Quote Modify
|
Hmm ... I am getting a flashback to the Fibonacci sequence ... those numbers ... so familiar ...
I am also getting an idea of how this puzzle was made up ...
Here's a hint (for those who don't like Googling):
Those numbers seem to differ by one, and they seem to be reciprocals ...
|
| « Last Edit: Oct 4th, 2002, 7:19am by James Fingas » |
IP Logged |
Doc, I'm addicted to advice! What should I do?
|
|
|
NickH
Senior Riddler
Gender:
Posts: 341
|
|
Re: NEW PUZZLE: Two real numbers
« Reply #3 on: Oct 4th, 2002, 2:11pm » |
Quote Modify
|
It's true the answer relates to the Fibonacci sequence, but really that's a coincidence. I made the puzzle up by considering the sum and product of the roots of a quadratic equation.
The starting equations can be manipulated into equations in x+y and xy. This greatly simplifies the algebra, and leaves us with a pair of quadratics.
Here's the solution...
(1) 1/x + 1/y = -1
(2) x^3 + y^3 = 4
(1) => x + y = -xy
(2) => (x + y)^3 - 3xy(x + y) = 4
=> -(xy)^3 + 3(xy)^2 - 4 = 0
By inspection, xy = -1 is a solution.
Hence (xy + 1)(xy - 2)^2 = 0.
So xy = -1, x + y = 1 or xy = 2, x + y = -2.
If xy = -1, x,y are roots of u^2 - u - 1 = 0.
Hence u = [1 +/- sqrt(5)]/2.
If xy = 2, x,y are roots of u^2 + 2u + 2 = 0.
This has complex roots: u = -1 +/- i.
Therefore, the real solutions are: (x,y) = ([1 +/- sqrt(5)]/2, [1 -/+ sqrt(5)]/2).
Nick
|
| « Last Edit: Oct 5th, 2002, 5:33am by NickH » |
IP Logged |
Nick's Mathematical Puzzles
|
|
|
Randy Williams
Guest
|
|
Re: NEW PUZZLE: Two real numbers
« Reply #4 on: Oct 4th, 2002, 2:38pm » |
Quote Modify
Remove
|
You can solve your sixth-order polynomial far enough by
factoring it into a fourth-order and a second-order
polynomial. I'll tell you the fourth-order factor, but
you'll have to work out the "golden ratios" from there:
x^4 + 4x^3 + 8x^2 + 8x + 4
|
|
IP Logged |
|
|
|
Benny
Uberpuzzler
Gender:
Posts: 1024
|
|
Re: Two real numbers
« Reply #5 on: Nov 21st, 2011, 5:55pm » |
Quote Modify
|
Does the system of eq.
(1) 1/x + 1/y = -1
(2) x^3 + y^3 = 4
gives us the Lucas numbers?
|
|
IP Logged |
If we want to understand our world — or how to change it — we must first understand the rational choices that shape it.
|
|
|
ThudnBlunder
wu::riddles Moderator
Uberpuzzler
The dewdrop slides into the shining Sea
Gender:
Posts: 4489
|
|
Re: Two real numbers
« Reply #6 on: Nov 22nd, 2011, 3:49am » |
Quote Modify
|
on Nov 21st, 2011, 5:55pm, BenVitale wrote:Does the system of eq.
(1) 1/x + 1/y = -1
(2) x^3 + y^3 = 4
gives us the Lucas numbers? |
|
Only the first two:
x1 + y1 = x2 + y2 = -2 = -L0 (complex roots)
x3 + y3 = x4 + y4 = 1 = L1 (real roots)
|
| « Last Edit: Nov 22nd, 2011, 9:56am by ThudnBlunder » |
IP Logged |
THE MEEK SHALL INHERIT THE EARTH.....................................................................er, if that's all right with the rest of you.
|
|
|
|
RIDDLES SITE
WRITE MATH!
Home 
Help 
Search 
Members 
Login 
Register
Reply
Notify of replies
Send Topic
Print