Two positive numbers x and 1-x add up to 1. Let the sum of their reciprocals be k. Find the only possible value of k such that x is a rational number. Hint: set up the equation and solve x with k in it. You'll encounter the quadratic formula in your solution.
so, did you follow the hint?
1/x + 1/(1-x) = k
kx(1-x) - 1 = 0
kx^2 - kx + 1 = 0
Now, you know the roots are
(k±√(k^2-4k))/2
For the roots to be rational, you must have
k^2-4k
a perfect square.
See what you can do with that.
Actually, the problem is misstated. You need k to be an integer. Otherwise, pick any positive rational number for x, and you have a different value for k.
1/3 + 2/3 = 1
3 + 3/2 = 11/2 = k
3/17 + 14/17 = 1
17/3 + 17/14 = k
But those k values are not integers.
To find the possible value of k, we need to solve the given equation and express it in terms of k.
Let's start by setting up the equation:
x + (1 - x) = 1
Simplifying this equation, we have:
1 = 1
This equation is true for any value of x. However, we're looking for the only possible value of k such that x is a rational number, which means the sum of the reciprocals of x and 1-x must yield a rational number.
Let's express the sum of the reciprocals of x and 1-x in terms of k:
1/x + 1/(1 - x) = k
To solve this equation, we need to find a common denominator and combine the fractions:
(1 - x + x) / (x(1 - x)) = k
1 / (x(1 - x)) = k
Since we want the value of x to be rational, we can assume x = p/q, where p and q are integers. Substituting this into the equation, we get:
1 / ((p/q)(1 - p/q)) = k
1 / ((p/q)(q - p)/q) = k
q / (p(q - p)) = k
Now, we have an equation involving q and p, which we can solve. Multiplying both sides by p(q - p), we get:
q = kp(q - p)
By expanding and rearranging the equation, we obtain:
kp^2 - (k - 1)pq + kp = 0
This equation is a quadratic equation in terms of p. We can apply the quadratic formula to find the values of p that satisfy this equation:
p = (-b ± √(b^2 - 4ac)) / 2a
where a = k, b = -(k - 1)q, and c = k.
To ensure that x is a rational number, the discriminant (b^2 - 4ac) must be a perfect square. So, let's simplify the discriminant:
b^2 - 4ac = (-(k - 1)q)^2 - 4(k)(kq) = (k^2 - 2k + 1)q^2 - 4k^2q
Simplifying further, we have:
(k^2 - 2k + 1 - 4k)q^2 = (k^2 - 6k + 1)q^2
For x to be rational, the discriminant must be a perfect square, which means (k^2 - 6k + 1)q^2 is a perfect square.
Now, let's consider the values of k that make (k^2 - 6k + 1) a perfect square. Factors of the form (k^2 - 6k + 1)(k^2 - 6k + 1) must also be a perfect square.
By expanding the product, we have:
(k^2 - 6k + 1)(k^2 - 6k + 1) = k^4 - 12k^3 + 37k^2 - 12k + 1
Let's denote this expression as M:
M = k^4 - 12k^3 + 37k^2 - 12k + 1
To find the possible values of k that yield a perfect square, we need to factorize M and determine its divisors. Then, we can substitute these divisors back into the equation q = kp(q - p) to find the corresponding values of q and p.
Since the possible values of k need to satisfy the quadratic equation and yield a rational number for x, the only way to find the specific value of k is by factoring M or evaluating the equation numerically.