# How do you find the value of the discriminant and determine the nature of the roots 2p^2+5p-4?

Jun 6, 2017

$\Delta = 57 > 0$ is not a perfect square, so the zeros are distinct, real and irrational.

#### Explanation:

Given:

$2 {p}^{2} + 5 p - 4$

Note that this is written in standard form:

$a {p}^{2} + b p + c$

with $a = 2$, $b = 5$ and $c = - 4$

The discriminant $\Delta$ is given by the formula:

$\Delta = {b}^{2} - 4 a c$

$\textcolor{w h i t e}{\Delta} = {\textcolor{b l u e}{5}}^{2} - 4 \left(\textcolor{b l u e}{2}\right) \left(\textcolor{b l u e}{- 4}\right)$

$\textcolor{w h i t e}{\Delta} = 25 + 32$

$\textcolor{w h i t e}{\Delta} = 57$

Since $\Delta > 0$ the given quadratic has two distinct real zeros. Since $\Delta = 57$ is not a perfect square, those zeros are irrational.

The zeros of $2 {p}^{2} + 5 p - 4$ are the roots of the equation:

$2 {p}^{2} + 5 p - 4 = 0$