Find k so that x^2 + 2(k − 4)x + 2k = 0 has equal (repeated) roots. Determine the value(s) of k.

Introduction / Context:
A quadratic has equal (repeated) real roots when its discriminant is zero. Apply this criterion to the parameterized quadratic x^2 + 2(k − 4)x + 2k = 0 to find all k that force equality of roots.

Given Data / Assumptions:

• Quadratic: x^2 + 2(k − 4)x + 2k = 0
• Equal roots ⇔ Discriminant D = 0

Concept / Approach:
Compute D = [2(k − 4)]^2 − 4*1*(2k). Set D = 0 and solve the resulting quadratic in k. Both solutions (if real) are acceptable parameter values.

Step-by-Step Solution:
D = 4(k − 4)^2 − 8k = 0Divide by 4: (k − 4)^2 − 2k = 0Expand: k^2 − 8k + 16 − 2k = 0 ⇒ k^2 − 10k + 16 = 0Solve: k = [10 ± √(100 − 64)]/2 = (10 ± 6)/2 ⇒ k = 8 or k = 2

Verification / Alternative check:
For k = 8: equation is x^2 + 8x + 16 = 0 ⇒ (x + 4)^2. For k = 2: x^2 − 4x + 4 = 0 ⇒ (x − 2)^2. Both have equal roots.

Why Other Options Are Wrong:
Pairs like “6 and 4”, “10 and 4”, “12 and 2” do not satisfy D = 0 for both k values; “Only 8” ignores k = 2, which also works.

Common Pitfalls:
Forgetting to divide by 4 when simplifying D or making sign errors when expanding (k − 4)^2. Always verify by completing the square.

Final Answer:
8 and 2