Question

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

k²x² – 2(2k – 1)x + 4 = 0

Answer 1

The given equation is k²x² - 2(2k - 1)x + 4 = 0

The given equation is in the form of ax² + bx + c = 0

Where a = k², b = -2(2k - 1) and c = 4

Therefore, the discriminant

D = b² - 4ac

= (-2(2k - 1))² - 4 × (k²) × (4)

= 4(2k - 1)² - 16k²

= 4(4k² + 1 - 4k) - 16k²

= 16k² + 4 - 16k - 16k²

= 4 - 16k

∵ Roots of the given equation are real and equal

∴ D = 0

⇒ 4 - 16k = 0

⇒ -16k = -4

`=> k=(-4)/-16`

⇒ `k = 1/4`

Hence, the value of k = `1/4`.