The Equation $2x^2 + (k-4)x + (k-4) = 0$where $k$ Is A Constant, Has Two Distinct Real Roots. Find The Set Of Possible Values Of $k$.

Introduction

In the realm of mathematics, particularly in algebra, equations play a crucial role in understanding various concepts and theorems. One such equation is the quadratic equation, which is of the form $ax^2 + bx + c = 0$. In this article, we will focus on the equation $2x^2 + (k-4)x + (k-4) = 0$, where $k$ is a constant. Our objective is to find the set of possible values of $k$ for which this equation has two distinct real roots.

The Nature of the Roots

To begin with, let's recall the nature of the roots of a quadratic equation. The roots of a quadratic equation $ax^2 + bx + c = 0$ can be real or complex, depending on the discriminant $b^2 - 4ac$. If the discriminant is positive, the equation has two distinct real roots. If the discriminant is zero, the equation has one real root. If the discriminant is negative, the equation has two complex roots.

The Discriminant of the Given Equation

Now, let's calculate the discriminant of the given equation $2x^2 + (k-4)x + (k-4) = 0$. The discriminant is given by the expression $(k-4)^2 - 4(2)(k-4)$. Simplifying this expression, we get $(k-4)^2 - 8(k-4)$.

Simplifying the Discriminant

To simplify the discriminant further, we can expand the square and combine like terms. Expanding the square, we get $k^2 - 8k + 16 - 8k + 32$. Combining like terms, we get $k^2 - 16k + 48$.

The Condition for Two Distinct Real Roots

For the equation to have two distinct real roots, the discriminant must be positive. Therefore, we must have $k^2 - 16k + 48 > 0$. This is the condition we need to satisfy to ensure that the equation has two distinct real roots.

Solving the Inequality

To solve the inequality $k^2 - 16k + 48 > 0$, we can factorize the quadratic expression. Factoring the expression, we get $(k-8)(k-6) > 0$. This inequality can be solved by considering the sign of the expression $(k-8)(k-6)$ for different ranges of $k$.

Solving the Inequality: Case 1

Let's consider the case when $k < 6$. In this case, both factors $(k-8)$ and $(k-6)$ are negative. Therefore, the product $(k-8)(k-6)$ is positive. Hence, the inequality $(k-8)(k-6) > 0$ is satisfied when $k < 6$.

Solving the Inequality: Case 2

Next, let's consider the case when $k > 8$. In this case, both factors $(k-8)$ and $(k-6)$ are positive. Therefore, the product $(k-8)(k-6)$ is positive. Hence, the inequality $(k-8)(k-6) > 0$ is satisfied when $k > 8$.

Solving the Inequality: Case 3

Finally, let's consider the case when $6 < k < 8$. In this case, the factor $(k-8)$ is negative, and the factor $(k-6)$ is positive. Therefore, the product $(k-8)(k-6)$ is negative. Hence, the inequality $(k-8)(k-6) > 0$ is not satisfied when $6 < k < 8$.

Conclusion

In conclusion, the set of possible values of $k$ for which the equation $2x^2 + (k-4)x + (k-4) = 0$ has two distinct real roots is given by the intervals $(-\infty, 6) \cup (8, \infty)$. This means that the value of $k$ can take any value less than 6 or greater than 8 for the equation to have two distinct real roots.

Final Answer

The final answer is $\boxed{(-\infty, 6) \cup (8, \infty)}$.