First, understand the question. We want to find the biggest integer (whole number) for a so that f(x) = ax^2 + 9x + 5 has two distinct, real zeros (it crosses the x-axis twice).
Several ways to approach this one.
1) I would begin by graphing the equation on Desmos.com. I tried several values for a (looking for the biggest one) and I found that the biggest integer value appears to be around 4.
2) Now, we need to prove that this is true by showing our work. But how?
We first, ask:
Q: "How can I find the roots or zeros of a quadratic equation?"
A: One way is by using the quadratic formula!
\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)
For our equation, the quadratic formula would be:
\(x = \frac{-9 \pm \sqrt{9^2 - 4(a)(5)}}{2(a)}\)
Now, remember that in order to have two real roots, the "discriminant" (the inside of the square root) MUST be positive. So what we're saying is that:
81 - 20 a > 0
But that's the same as:
81 > 20a
Or
4.05 > a
So, we are saying that in order to have two real roots, a must be less than 4.05. The biggest integer we can pick that is less than 4.05 is 4 !!! We are done! We have proven that a = 4.
