## 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*