Question

x²-2x+1 let a, b, c be positive integers such that the quadratic equation ax² - bx + c = 0 has two distinct roots in the interval (0,1). Find the smallest possible value of a.

Answer 1

The least value of a = 1

Explanation:

As it has two distinct roots . According to roll's theorem there should be a point where f'(x)=0

In a quadratic equation ax² + bx + c = 0 the point of maxima or minima is

x = - b/2a

We can find by differentiating it

2ax - b= 0

x = b/2a

So 0 < b/2a < 1

0 < b/a < 2

0 < b < 2a

a > b/2

then, the least value of b = 2 and the least value of a = 1