**Main scheme:**

Attempt to use discriminant $b^2 - 4ac$ (Need not be equated to zero) | M1 |

$144 - 4 \times k \times k = 0$ | A1 |

Attempt to solve for $k$ | M1 |

$k = 6$ | A1 | (4 marks)

**Total: 4 marks**

**Alternative for first 2 marks:**

Attempt to complete square $(x \pm p)^2 \pm q + c$, $p \neq 0$, $q \neq 0$ | M1 |

$1 - \frac{36}{k^2} = 0$ or equiv. | A1 |

**Other alternatives:**

(i) $x^2 + \frac{12}{k}x + 1$ must be equivalent to $(x+1)^2$ | M1 A1 |

Compare coefficients and attempt to solve for $k$: $\frac{12}{k} = 2$ $k = 6$ | M1 A1 |

(ii) Finding the root first, e.g. $(\sqrt{kx} + \sqrt{k})^2 = 0$, so $x = -1$ | M1 A1 |

Substitute the root to find $k$, $k = 6$ | M1 A1 |

**Note:** Answer only scores 2 marks: M0 A0 M1 A1. The first two marks would only be scored if solution then justifies that $k = 6$ gives equal roots.

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