# How do you find the value of the discriminant and state the type of solutions given 2p^2+5p-4=0?

##### 2 Answers
Apr 4, 2017

The solutions are $S = \left\{- 3.137 , 0.637\right\}$

#### Explanation:

Compare this equation to the quadratic equation

$a {x}^{2} + b x + c = 0$

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

The discriminant is

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

$= {\left(5\right)}^{2} - 4 \cdot 2 \cdot \left(- 4\right)$

$= 25 + 32$

$= 57$

As, $\Delta > 0$, there are 2 real solutions.

$x = \frac{- b \pm \sqrt{\Delta}}{2 a}$

The solutions are

${p}_{1} = \frac{- 5 + \sqrt{57}}{4} = 0.637$

${p}_{2} = \frac{- 5 - \sqrt{57}}{4} = - 3.137$

$\Delta = {5}^{2} - 4 \left(2\right) \left(- 4\right) = 25 + 32 = 57$ and so it'll be 2 real solutions.

#### Explanation:

One of the ways to see the discriminant is via the quadratic formula:

$x = \frac{- b \setminus \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$, and $a , b , c$ are substituted in when we have a trinomial in the form $a {x}^{2} + b x + c$

The discriminant is the ${b}^{2} - 4 a c$ part and is often identified with the Greek letter $\Delta$.

When $\Delta > 0$, you will end up with 2 real solutions.
When $\Delta = 0$, you will end up with 1 real solution.
When $\Delta < 0$, you will end up with 2 complex solutions.

Let's see what we get in our question:

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

${5}^{2} - 4 \left(2\right) \left(- 4\right) = 25 + 32 = 57$ and so it'll be 2 real solutions.

We can see that if we graph the equation:

graph{2x^2+5x-4 [-6.243, 6.243, -3.12, 3.123]}