(k-4)x^2+2(k-4)x+2=0

4 min read Jun 16, 2024

Solving the Quadratic Equation: (k-4)x^2 + 2(k-4)x + 2 = 0

This article will explore the solutions to the quadratic equation (k-4)x² + 2(k-4)x + 2 = 0, where k is a constant. We will delve into the different possibilities for the nature of the roots, considering the discriminant and its implications.

Understanding the Discriminant

The discriminant of a quadratic equation in the form ax² + bx + c = 0 is given by Δ = b² - 4ac. This value plays a crucial role in determining the nature of the roots:

Δ > 0: The quadratic equation has two distinct real roots.
Δ = 0: The quadratic equation has one real root (a double root).
Δ < 0: The quadratic equation has two complex roots (conjugate pairs).

Analyzing the Given Equation

In our case, we have a = (k-4), b = 2(k-4), and c = 2. Let's calculate the discriminant:

Δ = [2(k-4)]² - 4(k-4)(2)

Simplifying the expression:

Δ = 4(k-4)² - 8(k-4)

Δ = 4(k-4)(k-4 - 2)

Δ = 4(k-4)(k-6)

Determining the Nature of the Roots

Now, let's analyze the different possibilities based on the discriminant:

1. Δ > 0:

This implies that 4(k-4)(k-6) > 0. This inequality holds true when k < 4 or k > 6. In these cases, the quadratic equation has two distinct real roots.

2. Δ = 0:

This implies that 4(k-4)(k-6) = 0. This occurs when k = 4 or k = 6. In these cases, the quadratic equation has one real root (a double root).

3. Δ < 0:

This implies that 4(k-4)(k-6) < 0. This inequality holds true when 4 < k < 6. In these cases, the quadratic equation has two complex roots (conjugate pairs).

Conclusion

The quadratic equation (k-4)x² + 2(k-4)x + 2 = 0 will have different types of solutions depending on the value of k. The discriminant helps us determine the nature of these solutions:

Two distinct real roots: k < 4 or k > 6
One real root (a double root): k = 4 or k = 6
Two complex roots (conjugate pairs): 4 < k < 6