Solve Quadratic Equation: 3x²+12x+c=0

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SUMMARY

The discussion focuses on solving the quadratic equation 3x² + 12x + c = 0 by determining the values of c that yield different types of solutions. The quadratic formula x = (-b ± √(b² - 4ac)) / 2a is utilized, where the discriminant D = b² - 4ac plays a crucial role in identifying the nature of the solutions. For one real solution, the discriminant must equal zero; for two real solutions, it must be greater than zero; and for two nonreal (complex) solutions, it must be less than zero. Specifically, the discriminant for this equation simplifies to D = 12² - 4(3)c.

PREREQUISITES
  • Understanding of the quadratic formula
  • Knowledge of discriminants in quadratic equations
  • Basic algebraic manipulation skills
  • Familiarity with real and complex numbers
NEXT STEPS
  • Calculate the specific values of c for one real solution by setting the discriminant to zero.
  • Determine the range of c values that yield two real solutions by analyzing when the discriminant is positive.
  • Explore the conditions under which the discriminant is negative to find values of c that result in complex solutions.
  • Review additional examples of quadratic equations with varying coefficients to reinforce understanding of solution types.
USEFUL FOR

Students studying algebra, educators teaching quadratic equations, and anyone looking to deepen their understanding of the properties of quadratic functions.

PhantomTechnic
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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"
 
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Hello and welcome to MHB, PhantomTechnic! (Wave)

Let's review the quadratic formula:

Given the quadratic equation:

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

Then the solution is given by:

$$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$

Now, looking at that formula, what condition do we need for there to be only 1 solution?
 
MarkFL said:
Hello and welcome to MHB, PhantomTechnic! (Wave)

Let's review the quadratic formula:

Given the quadratic equation:

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

Then the solution is given by:

$$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$

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?
 
PhantomTechnic said:
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
 
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!
 

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