Question

find the value of k for which the quadratic equation (k-2)x^2 + 2(2k-3)x +(5k-6) = 0 has equal roots

Answer 1

Answer:

Step-by-step explanation:

Solution :-

(k - 2)x² + 2(2k - 3)x + (5k - 6) are real and equal.

Here, a = k - 2, b = 4k - 6, c = 5k - 6

D = b² - 4ac

= (4k - 6)²- 4 × (k - 2)(5k - 6)

= 16k² + 36 - 48k - 20k² + 64k - 48

= 4k² - 16k + 12

= k² - 4k + 3 = 0

= (k - 3)(k - 1) = 0

= k = 3, 1

Nature of roots of a quadratic equation:-

(i). If b² - 4ac > 0, the quadratic equation has two distinct real roots.

(ii). If b² - 4ac = 0,  the quadratic equation has two equal real roots.

(iii). If b² - 4ac < 0, the quadratic equation has no real roots.