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

Question

3105 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-06-06T15:11:16

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

$f(p) = 0$ 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]}

How do you solve 2p^2 - 7p + 3 > 02p^2 - 7p + 3 > 0?