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

Answer:

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:

#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]}