Find the two numbers whose product is 1 and whose sum is 1?

Answer 1

x + y = 1

xy = 1

y = 1 - x

x(1 - x) = 1

x - x^2 = 1

-x^2 + x - 1 = 0 or multiplying all terms by -1;

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

x^2 - x + 1 = 0

The roots are complex numbers. Use the quadratic formula and find them:

a = 1, b = -1, and c = 1

x = [-b + square root of (b^2 - (4)(a)(c)]/2a or

x = [-b - square root of (b^2 - (4)(a)(c)]/2a

So

x = [-(-1) + square root of ((-1)^2 - (4)(1)(1)]/2(1)

x = [1 + square root of (1 - 4]/2

x = [1 + square root of (- 3)]/2 or

x = [1 + square root of (-1 )(3)]/2; substitute (-1) = i^2;

x = [1 + square root of (i^2 )(3)]/2

x = [1 + (square root of 3)i]/2

x = 1/2 + [i(square root of 3]/2 and

x = 1/2 - [i(square root of 3)]/2

Since we have two values for x, we will find also two values for y

y = 1 - x

y = 1 - [1/2 + (i(square root of 3))/2]

y = 1 - 1/2 - [i(square root of 3)]/2

y = 1/2 - [i(square root of 3)]/2 and

y = 1 - [1/2 - (i(square root of 3))/2)]

y = 1 - 1/2 + [i(square root of 3))/2]

y = 1/2 + [i(square root of 3)]/2

Thus, these numbers are:

1. x = 1/2 + [i(square root of 3)]/2 and y = 1/2 - [i(square root of 3)]/2

2. x = 1/2 - [i(square root of 3)]/2 and y = 1/2 + [i(square root of 3)]/2

Let's check this:

x + y = 1

1/2 + [i(square root of 3)]/2 +1/2 - [i(square root of 3)]/2 = 1/2 + 1/2 = 1

xy = 1

[1/2 + [i(square root of 3)]/2] [1/2 - [i(square root of 3)]/2]

= (1/2)(1/2) -(1/2)[i(square root of 3)]/2] + [i(square root of 3)]/2](1/2) - [i(square root of 3)]/2] [i(square root of 3)]/2]

= 1/4 - [i(square root of 3)]/4 + [i(square root of 3)]/4 - (3i^2)/4; substitute ( i^2)=-1:

= 1/4 - [(3)(-1)]/4

= 1/4 + 3/4

= 4/4

=1

In the same way we check and two other values of x and y.