Bunuel
Two different positive numbers a and b each differ from their reciprocals by 1. What is a + b ?
(A) 1
(B) 2
(C) \(\sqrt{5}\)
(D) \(\sqrt{6}\)
(E) 3
It must be true that one number is more than its reciprocal, and the other is less than its reciprocal. We can let a be the one that is more than its reciprocal and hence b is the one that is less than its reciprocal. So we have:
a - 1/a = 1 and 1/b - b = 1
Subtracting these two equations, we have:
a - 1/a - 1/b + b = 0
a + b = 1/a + 1/b
a + b = (b + a)/(ab)
ab(a + b) = a + b
Since both a and b are positive, a + b ≠ 0. So we can divide both sides of the equation by a + b and obtain:
ab = 1
Now, let’s multiply the original first equation by a. We have:
a^2 - 1 = a
Similarly, let’s multiply the original second equation by b. We have:
1 - b^2 = b
Adding the two new equations, we have:
a^2 - b^2 = a + b
(a + b)(a - b) = a + b
Again, dividing both sides of the equation by a + b, we have:
a - b = 1
Squaring the above equation, we have:
a^2 - 2ab + b^2 = 1
Since ab = 1, we have:
a^2 - 2(1) + b^2 = 1
a^2 + b^2 = 3
Now add 2ab = 2 to the above equation, we have:
a^2 + 2ab + b^2 = 5
(a + b)^2 = 5
Taking the square root of both sides, we have:
a + b = √5
Answer: C