Equal roots condition (discriminant zero): Find all k such that x^2 − 2(1 + 3k)x + 7(3 + 2k) = 0 has equal (repeated) roots.

Introduction / Context:
A quadratic has equal (repeated) roots when its discriminant is zero. For ax^2 + bx + c = 0, the condition is b^2 − 4ac = 0. We apply this directly to the parameterized coefficients and solve the resulting quadratic equation in k.

Given Data / Assumptions:

• a = 1
• b = −2(1 + 3k)
• c = 7(3 + 2k)
• Equal roots ⇔ b^2 − 4ac = 0

Concept / Approach:
Compute b^2 and 4ac carefully, set their difference to zero, and simplify to a standard quadratic in k. Solve that equation to obtain all admissible k values.

Step-by-Step Solution:

Verification / Alternative check:
Substitute k = 2 and k = −10/9 back into D. In both cases, D becomes zero (by construction), confirming equal roots occur for exactly these values.

Why Other Options Are Wrong:

• Other sign combinations do not solve 9k^2 − 8k − 20 = 0.

Common Pitfalls:
Arithmetic slips when expanding (1 + 3k)^2 and in combining terms with the constant 84 and 56k. Keep coefficients organized.

Final Answer:
2, -10/9