The sum of reciprocals of 2 reals numbers is -1. The sum of their cubes is 4. What are the two numbers?
Can anyone explain how to solve this problem? Anything would be appreciated. Thank you.
2006-08-28 18:54:28 · 5 answers · asked by compstuft 2 in Education & Reference ➔ Homework Help
x^3 + y^3 = 4
1/x + 1/y = -1
apart from the symmetry, the system has 2 solutions: one complex
(x,y) = (i-1,-i-1) where i = sqrt(-1)
one real
(x,y) = ( (1-s)/(s-3), (1-s)/2 ) where s = sqrt(5)
2006-08-28 21:07:27 · answer #1 · answered by oracle 5 · 0⤊ 0⤋
I believe it is one number and that would be one, as the reciprocal is +1, if you add all sides having 1 then it would be 4.
2006-08-29 01:58:42 · answer #2 · answered by Esoteric 4 · 0⤊ 1⤋
i tried but the equation is too long
these will be the two equations:
a)1/x + 1/y = -1
b) x^3 + y^3 =4
u will get x^2/y^2=4/3
solve for this
mail me the answer if u get the two no.s
2006-08-29 03:12:07 · answer #3 · answered by Anonymous · 0⤊ 1⤋
x/y + y/x = -1
(x/y)^2 + (y/x)^2 = 4
solve for x in the first equation, get rid of the denominator:
i don't really remember the principles, but i hope this helps u for now. good luck
2006-08-29 02:48:33 · answer #4 · answered by sasmallworld 6 · 0⤊ 1⤋
ask your teacher
2006-08-29 02:00:16 · answer #5 · answered by Anonymous · 0⤊ 1⤋