x2-6(p%2B1)x%2B3(p%2B9)%3D0.jpeg "(p+1)x2-6(p+1)x+3(p+9)=0")
Solving the Quadratic Equation: (p+1)x² - 6(p+1)x + 3(p+9) = 0
This article explores the solution to the quadratic equation (p+1)x² - 6(p+1)x + 3(p+9) = 0, where 'p' is a constant. We will utilize the quadratic formula and explore the nature of the roots depending on the value of 'p'.
Understanding the Quadratic Formula
The quadratic formula is a powerful tool used to solve equations of the form ax² + bx + c = 0. It states that the solutions (roots) of this equation are given by:
x = [-b ± √(b² - 4ac)] / 2a
Applying the Formula to our Equation
In our case, we have:
- a = (p+1)
- b = -6(p+1)
- c = 3(p+9)
Substituting these values into the quadratic formula, we get:
x = [6(p+1) ± √((-6(p+1))² - 4(p+1)(3(p+9)))] / 2(p+1)
Simplifying the Expression
Let's simplify the expression inside the square root:
- (-6(p+1))² = 36(p+1)²
- 4(p+1)(3(p+9)) = 12(p+1)(p+9)
Therefore, the expression under the square root becomes:
36(p+1)² - 12(p+1)(p+9) = 12(p+1)(3(p+1) - (p+9)) = 12(p+1)(2p-6) = 24(p+1)(p-3)
Substituting this back into our solution for x, we get:
x = [6(p+1) ± √(24(p+1)(p-3))] / 2(p+1)
Analyzing the Roots
The nature of the roots depends on the discriminant, which is the expression under the square root:
- Discriminant > 0: Two distinct real roots
- Discriminant = 0: One real root (a double root)
- Discriminant < 0: Two complex roots
In our case, the discriminant is 24(p+1)(p-3). Let's analyze its sign for different values of 'p':
- p < -1: Both (p+1) and (p-3) are negative, making the discriminant positive. This leads to two distinct real roots.
- -1 < p < 3: (p+1) is positive, while (p-3) is negative, making the discriminant negative. This results in two complex roots.
- p > 3: Both (p+1) and (p-3) are positive, making the discriminant positive. This again leads to two distinct real roots.
Conclusion
The solution to the quadratic equation (p+1)x² - 6(p+1)x + 3(p+9) = 0 can be obtained using the quadratic formula. The nature of the roots depends on the value of 'p'. We have explored the different scenarios and determined that the equation will have two distinct real roots for p < -1 or p > 3, two complex roots for -1 < p < 3, and one real root (double root) for p = -1 or p = 3.