Equivalently, we need to find an integer a such that F(x) = √(5x − 1)−6+x^2

has at least one root in the interval [a, a + 1].

Notice that F(1) = √4 − 6 + 1 = −3 < 0 , while F(2) = √9 − 6 + 4 = 1 > 0 ,and F(x) is continuous on the interval [1, 2]. Therefore, by the Intermediate Value

Theorem, there is some c such that 1 ≤ c ≤ 2 and F(c) = 0. Therefore, √5c − 1 − 6 + c^2 = 0, so √5c − 1 = 6 − c^2, or in other words the equation

has at least one solution in the interval [1, 2].

So a = 1.