Question

Find the values of k for which the following equation has equal roots:

(3k + 1)x² + 2(k + 1)x + k = 0

Answer 1

Given: (3k + 1)x² + 2(k + 1)x + k = 0

Step-wise calculation:

1. Compare with ax² + bx + c = 0:

a = 3k + 1, b = 2(k + 1), c = k

2. For equal (repeated) roots the discriminant D = b² – 4ac must be 0.

D = [2(k + 1)]² – 4(3k + 1)(k) 

= 4(k + 1)² – 4k(3k + 1)

3. Divide by 4:

(k + 1)² – k(3k + 1) = 0

k² + 2k + 1 – 3k² – k = 0

–2k² + k + 1 = 0 

Multiply by –1: 

2k² – k – 1 = 0

4. Solve quadratic in k:

Discriminant Δ = (–1)² – 4 × 2 × (–1) 

= 1 + 8

= 9

`k = (1 + 3)/4 = 1` or `k = (1 - 3)/4 = -1/2`.

Also note we require a ≠ 0 for a quadratic, i.e. `k ≠ -1/3`. 

Neither solution equals `−1/3`, so both are valid.

k = 1 or k = `−1/2`.