(k-12)x2+2(k-12)x+2=0

4 min read Jun 16, 2024

Solving the Quadratic Equation: (k-12)x² + 2(k-12)x + 2 = 0

This article will guide you through solving the quadratic equation (k-12)x² + 2(k-12)x + 2 = 0. We'll explore different methods and analyze the conditions for real solutions.

Understanding the Quadratic Formula

The quadratic formula is a powerful tool used to solve any quadratic equation of the form ax² + bx + c = 0. It states that:

x = (-b ± √(b² - 4ac)) / 2a

where:

a, b, and c are the coefficients of the quadratic equation.

Applying the Quadratic Formula

Let's apply the quadratic formula to our equation:

a = (k-12)
b = 2(k-12)
c = 2

Substituting these values into the quadratic formula, we get:

x = (-2(k-12) ± √((2(k-12))² - 4(k-12)(2))) / 2(k-12)

Simplifying the Expression

Let's simplify the expression:

x = (-2(k-12) ± √(4(k-12)² - 8(k-12))) / 2(k-12)

x = (-2(k-12) ± √(4(k-12)(k-12-2))) / 2(k-12)

x = (-2(k-12) ± √(4(k-12)(k-14))) / 2(k-12)

x = (-2(k-12) ± 2√((k-12)(k-14))) / 2(k-12)

x = (- (k-12) ± √((k-12)(k-14))) / (k-12)

Analyzing the Solutions

The solutions to this quadratic equation depend on the value of k:

Case 1: (k-12)(k-14) > 0
- In this case, the discriminant (the expression under the square root) is positive, resulting in two distinct real solutions.
- The solutions will be real and distinct.
Case 2: (k-12)(k-14) = 0
- This results in a double root, meaning there is only one solution.
- The solution will be real and repeated.
Case 3: (k-12)(k-14) < 0
- In this case, the discriminant is negative, resulting in no real solutions.
- The solutions will be complex conjugates.

Conclusion

By applying the quadratic formula and analyzing the discriminant, we've determined the conditions for real solutions to the equation (k-12)x² + 2(k-12)x + 2 = 0. Remember to consider the different cases based on the value of k to fully understand the nature of the solutions.