Question

(k-5)xsquare+2(k-5)x=2 have equal roots find the value of k

Answer 1

Given equation :-

( k - 5 )x² + 2 ( k - 5 )x = 2

( k - 5 )x² + 2 ( k - 5 )x - 2 = 0


a = ( k - 5 )

b = 2 ( k - 5 )
= 2k - 10

c = 2


As we know that for equal roots the value of discriminate is equal to zero.

So,

D = b² - 4ac

D = ( 2k - 10 )² - 4 × ( k - 5 ) × 2

0 = 4k² + 100 - 40k - 8 ( k - 5 )

0 = 4k² + 100 - 40k - 8k + 40

0 = 4k² + 140 - 48k

( dividing by 4 )

0 = k² + 35 - 12k

0 = k² - 12k + 35

0 = k² - 7k - 5k + 35

0 = k ( k - 7 ) - 5 ( k - 7 )

0 = ( k - 5 ) ( k - 7 )


* ( k - 5 ) = 0
k = 5


* ( k - 7 ) = 0
k = 7


For the value of k 5 and 7 both are applicable!