A positive real number is 1 more than its reciprocal. Find the number:
A. [1-sqrt(5)]/2
B. 2
C. [1+sqrt(5)]/2
D. No solution
A positive real number is 1 more than its reciprocal. Find the number:
A. [1-sqrt(5)]/2
B. 2
C. [1+sqrt(5)]/2
D. No solution
N = 1/N + 1 or N^2 - N - 1 = 0
From this derives N = (1 + sqrt5)/2
1.6
To find the positive real number that is 1 more than its reciprocal, let's first set up the equation:
Let x be the positive real number.
The reciprocal of x is 1/x.
According to the problem, x is 1 more than its reciprocal, so we have the equation:
x = 1 + 1/x
To solve this equation, we can start by multiplying both sides of the equation by x:
x^2 = x + 1
Next, rearrange the equation to bring all terms to one side:
x^2 - x - 1 = 0
Now, we have a quadratic equation. To solve it, we can use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
In this case, a = 1, b = -1, and c = -1. Plugging these values into the quadratic formula, we get:
x = (-(-1) ± √((-1)^2 - 4(1)(-1))) / (2(1))
x = (1 ± √(1 + 4)) / 2
x = (1 ± √5) / 2
So, the solutions for x are:
x = (1 + √5) / 2 (Option C)
x = (1 - √5) / 2 (Option A)
Since we are looking for a positive real number, we can see that the only valid solution is:
x = (1 + √5) / 2 (Option C)
Therefore, the number that satisfies the given condition is option C: [1 + sqrt(5)]/2.