Question

Find the values of k for which the roots are real and equal in each of the following 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 x (k²) x (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

`rARrk=(-4)/-16`

⇒ k = 1/4

Hence, the value of K = 1/4