Question

help and solve plz number 16 ok fast please

Answer 1

Given :-
Equation :-
( k - 2 )x² + 2( 2k - 3 )x + ( 5k - 6 ) = 0


Here,

a = ( k - 2 )

b = 2 ( 2k - 3 )

c = ( 5k - 6 )


As we know that

=> Discriminate ( D ) of quadratic equation will be Zero when the roots of equation are equal !!

So,

=> D = b² - 4ac

0 = [ 2 ( 2k - 3 ) ]² - 4 × ( k - 2 ) × ( 5k - 6 )

0 = [ 4k - 6 ]² - ( 4k - 8 ) ( 5k - 6 )

0 = 16k² + 36 - 48k - ( 20k² - 24k -40k+48)

0 = 16k² + 36 - 48k - 20k² + 24k + 40k- 48

0 = - 4k² + 64k - 48k - 12

0 = - 4k² + 16k - 12

( dividing by 4 )

0 = - k² + 4k - 3

k² - 4k + 3 = 0

k² - 3k - k + 3 = 0

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

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

* ( k - 1 ) = 0
k = 1

* ( k - 3 ) = 0
k = 3