Question

There are two numbers, their product as well as sum is 1, find the numbers.....
(no spam)(legit answers)

Answer 1

let the numbers be x and y

given that x+y = 1

=> y = 1-x

Also, xy = 1

=> x(1-x) = 1

=> x - x^2 = 1

=> x^2 - x + 1 = 0

Solve the quadratic equation to get the values.

$x = \frac{ -1 \pm \sqrt{ {( - 1) }^{2} - 4 \times 1 \times 1} }{ \sqrt{2 \times 1} } \\ \\ x = \frac{ - 1\pm \sqrt{ 1 - 4}}{2} \\ \\ x = \frac{ - 1\pm i \sqrt{3}}{2}$

So there is no real number which satisfies the conditions.

Above two values of x are the complex solutions.

Answer 2

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let the numbers be x and y

given that x+y = 1

=> y = 1-x

Also, xy = 1

=> x(1-x) = 1

=> x - x^2 = 1

=> x^2 - x + 1 = 0

Solve the quadratic equation to get the values.

$x = \frac{ -1 \pm \sqrt{ {( - 1) }^{2} - 4 \times 1 \times 1} }{ \sqrt{2 \times 1} } \\ \\ x = \frac{ - 1\pm \sqrt{ 1 - 4}}{2} \\ \\ x = \frac{ - 1\pm i \sqrt{3}}{2}$

So there is no real number which satisfies the conditions.

Above two values of x are the complex solutions.

HOPE SO IT WILL HELP.........