How do you find the value of the discriminant and state the type of solutions given 2p^2+5p-4=02p2+5p−4=0?

2 Answers
Narad T.
Apr 4, 2017

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

Explanation:

Compare this equation to the quadratic equation

ax^2+bx+c=0ax2+bx+c=0

2p^2+5p-4=02p2+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

MathFact-orials.blogspot.com
Apr 4, 2017

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]}

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