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

4 min read Jun 16, 2024
(k-12)x2+2(k-12)x+2=0

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.

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