Question

Find value of K which given equation has real & equal roots
k^2X^2-2(k-1)X+4​

Answer 1

Hope this is helpful...

Answer 2

Answer:

k =1/2 = 0.5

Step-by-step explanation:

k²x² - 2kx + 2x +4

Since it has two real root and equal root

therefor b² - 4ac should be 0

here,

a = k²

b = -2

c = 4

now, substituting this value in b² - 4ac =0

(-2)² - 4 × k² × 4 = 0

⇒ 4 - 16k² = 0

⇒ -16k² = -4

⇒ 16k² = 4

⇒ k² = 4/16

⇒ k = √(4/16) = √(2 × 2/ 4 × 4)

⇒ k = 2/4

⇒ k =1/2 = 0.5