Question

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

(k + 1)x² - 2(k - 1)x + 1 = 0

Answer 1

The given equation is (k + 1)x² - 2(k - 1)x + 1 = 0

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

where a = (k + 1), b = -2(k - 1) and c = 1

Therefore, the discriminant

D = b² - 4ac

= (-2(k - 1))² - 4 x (k + 1) x (1)

= 4(k - 1)² - -4k - 4

= 4(k² + 1 - 2k) - 4k - 4

= 4k² + 4 - 8k - 4k - 4

= 4k² - 12k

∵ Roots of the given equation are real and equal

∴ D = 0

4k² - 12k = 0

4k(k - 3) = 0

k = 0

Or k - 3 = 0

k = 3

Hence, the value of k = 0, 3.