find 2 positive real #'s that differ by 1 and have a product of 1?

Answer 1

So, x-y = 1 and x*y=1. Solve the first for x, substitute into the second and solve, and substitute into the first. Then solve for y, and use that information to get x.

For, suppose the sum was 11 and the product was 24:

(1) x+y = 11. [Given.]
(2) x*y = 24. [Given.]

Isolating x in (1),

(3) x = 11-y.

Substituting 11-y for x in (2),

(4) y*(11-y) = 24;
11*y - y^2 = 24;
0 = y^2 - 11*y + 24.

y = [11 +/- sqrt( (-11)^2 - 4*1*24 )]/2*1
= [11 +/- sqrt( 121 - 96 )]/2
= [11 +/- sqrt( 25 )]/2 = [11 +/- 5]/2
= 16/2 or 6/2
= 8 or 3.

So x = 3 or 8, by either (1) or (2).

So the solution to this (similar) problem is 3 and 8.

Good enough?

Answer 2

x * y = 1
y = x + 1

x * ( x + 1 ) = 1
x^2 + x = 1
x^2 + x -1 = 0

Quadratic formula:
[-1 +/- sqrt(1^2 - 4*1*-1)] / 2*1
[-1 +/- sqrt(5)]/2

x= -(sqrt(5) + 1)/2 = -1.61803...
y= -(sqrt(5) -1)/2 = -0.618034...

OR

x= (sqrt(5) -1)/2 = 0.618034...
y= (sqrt(5) + 1)/2 = 1.61803...

The second answer of "y" is also known as "phi" or the Golden Ratio by which most things in nature share (according to some).

Try measuring your height, and then measure your height from the ground to your belly button. Take the first measurement and divide it by the second, and you should have a number close to 1.6! It's a fun little experiment. You can also try it with your arms and legs and their respective joints!

Answer 3

Set up 2 equations: x - y = 1 and xy = 1

By substitution, you will come to x^2 -x-1 = 0

Use the quadratic formula to solve this, put the answers back into eq #1, and you will have your answers.

The roots are the ratio of 2 adjacent terms of the Fibonacci series when expanded infinitely.

Ans: 1.618.... and .618....... or -1.618... and -.618....

Answer 4

x-y=1
xy=1

x=1+y

(1+y)y=1
y + y[2] =1

ok, take it from there.. I give up :-)

Answer 5

I don't think there are any but hey...i could be wrong.