Solve Quadratic Equation: 3x²+12x+c=0

[PhantomTechnic]

Hello I've been stuck on this test review question for a few days, and I can't figure it out. Can someone help out?
"3x²+12x+c=0, Find solutions for c, where there is 1 real solution, 2 real solutions, and 2 nonreal(complex) solutions"

 

[MarkFL]

Hello and welcome to MHB, PhantomTechnic! (Wave)

Let's review the quadratic formula:

Given the quadratic equation:

\(\displaystyle ax^2+bx+c=0\)

Then the solution is given by:

\(\displaystyle x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\)

Now, looking at that formula, what condition do we need for there to be only 1 solution?

 

[PhantomTechnic]

Hello and welcome to MHB, PhantomTechnic! (Wave)

Let's review the quadratic formula:

Given the quadratic equation:

\(\displaystyle ax^2+bx+c=0\)

Then the solution is given by:

\(\displaystyle x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\)

Now, looking at that formula, what condition do we need for there to be only 1 solution?

Wouldn't it be when it equals zero?

 

[MarkFL]

Wouldn't it be when it equals zero?

When what equals zero? You are headed in the right direction, but I want to make certain you are talking about the correct expression...D

 

[suluclac]

To answer Mark, D = 12² - 4(3)c, where D is the discriminant. What equals 0?
Solve for c. What do you get? It's probably not a prime number!