Question

Find the values of k for which the roots are real and equal in the following quadratic equation:

4x² – 2(k + 1)x + (k + 1) = 0

Answer 1

Given: 4x² – 2(k + 1)x + (k + 1) = 0

Step-wise calculation:

1. Compare with ax² + bx + c = 0:

a = 4, b = –2(k + 1), c = (k + 1)

2. Discriminant D = b² – 4ac. For equal (repeated) real roots we require D = 0 (discriminant test).

3. Compute D: D = [–2(k + 1)]² – 4 × 4 × (k + 1) 

= 4(k + 1)² – 16(k + 1) 

= 4(k + 1)[(k + 1) – 4]

= 4(k + 1)(k – 3)

4. Set D = 0: 4(k + 1)(k – 3) = 0

⇒ k + 1 = 0 or k – 3 = 0

⇒ k = –1 or k = 3

5. Check roots: If k = –1: equation → 4x² = 0

⇒ Double root x = 0

If k = 3: equation → 4x² – 8x + 4 = 0

⇒ (x – 1)² = 0

⇒ Double root x = 1

The values of k for which the roots are real and equal are k = –1 and k = 3.