Question

Find K for which equation have real and equal roots (K+1)x^2-2(k-1)x+1​

Answer 1

Answer:

answer for the given problem is given

Answer 2

⬛️ As the roots are real and equal

⇒ $\therefore{D = 0}$

we know,

d = b² - 4ac

so here,

➡️b = 2 ( k + 1)

➡️a = ( k + 1 )

➡️c = 1

now,

⇒ D = 0

⇒ [2(k+ 1)]² - 4 × ( k + 1) ×1 = 0

⇒[4 ( k² + 2k + 1)]- 4k- 4 = 0

⇒4k² + 8k+ 4 - 4k -4 = 0

⇒4k² + 4k = 0

⇒ 4k(k + 1 ) = 0

so , either -

⇒ 4k(k + 1 ) = 0

⇒ 4k = 0/(k+1)

⇒4k = 0

⇒k = 0/4

⇒k = 0

or,

⇒4k(k+1) = 0

⇒(k + 1 ) = 0/4k

⇒k + 1 = 0

⇒k = -1

Things to Know:-

D = b² - 4ac is called Discriminant .

Nature of the Roots depend on "D"

⬛️ If roots are real and distinct then,

⇒ D >0

⬛️ If roots are real and equal then,

⇒ D = 0

⬛️ If roots are not real or imaginary then,

⇒ D <0