Question

Find 'K' if the equation kx( x-2)+6=0 has real and equal root​

Answer 1

Answer:

here is the answer....

Answer 2

=> kx (x-2) + 6 = 0

=> kx² - 2kx + 6 = 0

on comparing this with ax² + bx + c

we get.....

a = k ; b = -2k ; c = 6

For the roots to be real nd equal....

thn,

D = 0

=> b² - 4ac = 0

=> (-2k)² - 4 (k) (6) = 0

=> 4k² - 24k = 0

=> 4k ( k - 6 ) = 0

now....

4k = 0

=> k = 0/4

=> k = 0

k - 6 = 0

=> k = 6

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