Quadratic Equation- Parabola Intersection

Question

The point (1,2) is on graph of the quadratic function f(x)= ax^2+bx+1. Determine the values of a, such that the graph of f(x) intersects the x-axis at two distinct point.

Answer is : a > 3 + 2 √ __ 2 and a < 3 – 2 √ __ 2 , a ≠ 0

but not sure how to get it

Answer 1

The idea is to first find the values of a that result in an intersection at ONE point (the vertex). This occurs when the discriminant (b² - 4ac) equals zero. In this case, that becomes b² - 4a = 0

From the point that was given, 2 = a1² + b1 + 1

This gives us 1 = a + b

Solve simultaneously by substituting 1-a in for b:

(1-a)² - 4a = 0

1 - 2a + a² - 4a = 0

a² - 6a + 1 = 0

Using the quadratic formula, we find that a = 3 ± 2√2

For a quadratic to have 2 real solutions, the discriminant must be greater than 0. We need to test a value that above or below the "one solution" value to see which way the inequality must go.

Starting with a = 3 + 2√2, let's try a = 6, the smallest integer above 3 + 2√2. It gives a positive discriminant, so we know that a > 3 + 2√2 works.

Going the other "one solution" value (3 - 2√2), let's check out a = 0, the largest integer below it. This results in a positive discriminant, so we also know that a < 3 - 2√2 works.