(a)

\( k = 1 \), \( k = 9 \) (M1A1M1A1A1, N3).

Working:
Given: \( k x^2 + (k – 3)x + 1 = 0 \) has two equal real roots.
Condition: Discriminant \( \Delta = 0 \).
Discriminant: \( \Delta = (k – 3)^2 – 4 \times k \times 1 \).
Substitute: \( (k – 3)^2 – 4 \times k \times 1 = k^2 – 6k + 9 – 4k = k^2 – 10k + 9 \).
Set \( \Delta = 0 \): \( k^2 – 10k + 9 = 0 \).
Solve: \( (k – 1)(k – 9) = 0 \).
Solutions: \( k = 1 \), \( k = 9 \). (M1 for discriminant attempt, A1 for correct substitution, M1 for setting discriminant to zero, A1 for quadratic, A1 for solutions)

[5 marks]

(b)

\( k = 1 \), \( k = 9 \) (A2, N2).

Working:
Given: \( x^2 + (k – 3)x + k = 0 \) has two equal real roots.
Discriminant: \( \Delta = (k – 3)^2 – 4 \times 1 \times k \).
Substitute: \( (k – 3)^2 – 4k = k^2 – 6k + 9 – 4k = k^2 – 10k + 9 \).
Set \( \Delta = 0 \): \( k^2 – 10k + 9 = 0 \).
Solve: \( (k – 1)(k – 9) = 0 \).
Solutions: \( k = 1 \), \( k = 9 \). (A2 for correct solutions)

[2 marks]

Total [7 marks]