The sum of reciprocals of two real numbers is -1. The sum of their cubes is 4. Find the numbers. *HELP ! );?

Question

The answers are not whole numbers !

Answer 1

let x = 1st number
let y = 2nd number

** Okay, they tell you that "the sum of the reciprocals of two real number is -1.... So this is how you write that statement in an algebraic equation **

1 / x + 1 / y = -1

** .... but -1 is the same as saying -y/y, right?? So... **

1 / x + 1 / y = -y/y

1 / x = -y/y - 1/y

1 / x = (-y -1) / y

x = - y / (y+1) ... so this is what "x" is in terms of "y"...


** Okay, they also tell you that "the sum of the cubes of the two real number is 4.... So this is how you write that statement in an algebraic equation **....

x^3 + y ^3 = 4

** NOW.... remember you had found "x" in terms of "y"??? Well, "cube" that expression and then substitute that expression for x^3 in the "x^3 + y^3 = 4" equation... like this..."

x = - y / (y+1) .... cubing this you get... x^3 = - y^3 / (y+1)^3

So.... when you substitute for x^3...

"x^3 + y^3 = 4" becomes...

- y^3 / (y+1)^3 + y^3 = 4

** multiply the whole above equation by (y+1)^3 **

- y^3 + y^3 (y+1)^3 = 4(y+1)^3

** now expand the cubes **

-y^3 + y^3(y^2 + 2y + 1)(y+1) = 4(y^2 + 2y + 1)(y+1)

** now it's just algebra from here on... **

-y^3 + y^3(y^3 + 2y^2 + y + y^2 + 2y + 1) = 4(y^3 + 2y^2 + y + y^2 + 2y + 1)

-y^3 + y^3(y^3 + 3y^2 + 3y + 1) = 4(y^3 + 3y^2 + 3y + 1)

-y^3 + y^6 + 3y^5 + 3y^4 + y^3 = 4y^3 + 12y^2 + 12y + 4

y^6 + 3y^5 + 3y^4 = 4y^3 + 12y^2 + 12y + 4 ... This is what you should finally get...

By trial and error (i.e., picking random values of "y"), I found that the y-value that makes this equation true is.... x = = -0.618034 ... because then the left side = 0.2229123.... and the right side = 0.2229123

Now that you know that one of the numbers is -0.618034, you can use the fact that "the sum of the reciprocals of two real number is -1" to find the other real number....

1 / x + 1 / y = -1

now plug in x = -0.618034... and find "y"

1/(-0.618034) + 1/y = -1

1/y = -1 - [1/(-0.618034)]

1/y = -1 - [-1.618034]

1/y = -1 +1.618034

1/y = 0.618034

y = 1/0.618034

y = 1.618034

So our two real numbers are -0.618034 and 1.618034

NOW.... to check.... use the fact that "the sum of their cubes is 4"....

x^3 + y ^3 = 4

Substitute the two real numbers and see if the sum of their cubes really does equal "4"....

So....

x^3 + y ^3 = 4

(-0.618034)^3 + (1.618034)^3 = 4 .... Is this true???

-0.2360679 + 4.2360681 = 4 ???

4 = 4 YES!!!

So our two real numbers are -0.618034 and 1.618034 ANSWER

.....because....

1) The sum of the reciprocals is -1.

1/(-0.618034) + 1/(1.618034) = -1

AND

2) The sum of their cubes is 4.

(-0.618034)^3 + (1.618034)^3 = 4


HOPE THIS HELPS!!! I give you a yellow star for asking a most interesting and challenging question that made me think!!! =)

Answer 2

I did it and got whole numbers. Are you sure?