Question

How do you solve `2p^2 - 7p + 3 > 0`?

Answer 1

`f(p) = 2p^2-7p+3 = (2p-1)(p-3)`

`f(p) = 0` when `p=1/2` or `p=3`
So the graph of `f(p)` crosses the horizontal axis at `(1/2, 0)` and `(3, 0)`.

For large positive or negative values of `p`, the `2p^2` term will dominate, resulting in `f(p) > 0`

Hence `f(p) > 0` when `p < 1/2` or `p > 3`

graph{2x^2-7x+3 [-10, 10, -5, 5]}