Find two positive real numbers that differ by 1 and have a product of 1.
Define your numbers first:
1st number = x
2nd number = x +1
Problem says "Product of 1", so multiply
x(x+1)=1
x^2+x=1
x^2+x-1=0
From here use Quadratic Formula or complete the square to solve.
To find two positive real numbers that differ by 1 and have a product of 1, we can set up a system of equations.
Let's assume the two numbers are x and y.
According to the given information, the two numbers differ by 1, so we can write the equation as:
x - y = 1 ----(1)
Also, the product of the two numbers is 1, so we can write the equation as:
xy = 1 ----(2)
To solve this system of equations, we can solve equation (1) for x and substitute it into equation (2).
From equation (1), we can add y to both sides:
x = y + 1
Substituting this value of x into equation (2), we get:
(y + 1)y = 1
Expanding the equation, we have:
y^2 + y = 1
Rearranging the equation, we get:
y^2 + y - 1 = 0
This is a quadratic equation. By using the quadratic formula, we can solve for y:
y = (-b ± √(b^2 - 4ac)) / (2a)
In this case, the quadratic equation is in the form of: ay^2 + by + c = 0, where a = 1, b = 1, and c = -1.
Now, substituting the values of a, b, and c into the quadratic formula, we get:
y = (-(1) ± √((1)^2 - 4(1)(-1))) / (2(1))
Simplifying further:
y = (-1 ± √(1 + 4)) / 2
y = (-1 ± √5) / 2
So, the two possible values of y are:
y₁ = (-1 + √5) / 2
y₂ = (-1 - √5) / 2
To find the corresponding values of x, we can substitute these values of y back into equation (1):
For y₁ = (-1 + √5) / 2:
x₁ = y₁ + 1 = (-1 + √5) / 2 + 1 = (1 + √5) / 2
For y₂ = (-1 - √5) / 2:
x₂ = y₂ + 1 = (-1 - √5) / 2 + 1 = (1 - √5) / 2
Therefore, the two positive real numbers that differ by 1 and have a product of 1 are:
(1 + √5) / 2 and (1 - √5) / 2