Question

( k +1) x2 - 4kx +9 =0 have real and equal roots find x

Answer 1

Answer:

Step-by-step explanation:

We have the equation in form ax² + bx + c

i.e  (k+1) x² - 4kx + 9 = 0 (where a = k+1 ; b = -4k ; c = 9 )

if the quadratic equation has equal and real roots then,

D = 0

b² - 4ac = 0

(-4k)² - 4 (k+1) (9) = 0

16k² - 36k +36 = 0

4 (4k² - 9k + 9) = 0 (taking 4 common)

therefore, 4k² - 9k + 9 = 0

solving the equation ,

(Factorizing) 4k² - 12k - 3k + 9 = 0

4k (k-3) - 3 (k - 3) + 0

(k-3)(4k-3) = 0

thus, k = 3 or 3/4

PUT THESE VALUES  OF "K" ONE BY ONE IN EQUATION AND THEREBY GET VALUE OF X.

HOPE IT HELPS  :))

 

                             

Answer 2

Answer:

b² - 4ac = 0 [because roots are real and equal]

(-4k)^2 - 4(k+1)(9) = 0

16k^2 - 36k +36 = 0

4k^2 - 9k + 9 = 0 (dividing throughout by 4)

4k^2 - 9k + 9 = 0

(k-3)(4k-3) = 0

thus, k = 3 or k = 0.75