Question

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

Answer 1

The solutions are `S={-3.137, 0.637}`

Explanation:

Compare this equation to the quadratic equation

`ax^2+bx+c=0`

`2p^2+5p-4=0`

The discriminant is

`Delta=b^2-4ac`

`=(5)^2-4*2*(-4)`

`=25+32`

`=57`

As, `Delta>0`, there are 2 real solutions.

`x=(-b+-sqrtDelta)/(2a)`

The solutions are

`p_1=(-5+sqrt57)/4=0.637`

`p_2=(-5-sqrt57)/4=-3.137`

Answer 2

`Delta=5^2-4(2)(-4)=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 = (-b \pm sqrt(b^2-4ac)) / (2a)`, and `a, b, c` are substituted in when we have a trinomial in the form `ax^2+bx+c`

The discriminant is the `b^2-4ac` 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(2)(-4)=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]}