Question

Find the values of ‘k’, for which the quadratic equation kx2− (8k + 4) + 81 = 0 has real and equal

roots?​

Answer 1

Step-by-step explanation:

GIVEN

Kx²-(8k+4)+81

Kx²-8k-4+81

kx²-8k+77

compare the eq with ax²+bx+c

we get a = k, b= 0( as there is no which is in the form of bx) ,c= -8k+77

if the eq has real and equal roots we know that discriminant D= 0

D=b²-4ac

0= 0²-4×k×(-8k+77)

0= - 4k(-8k+77)

so -4k=0

and -8k+77=0

so k = -77/-8

k=77/8

Answer 2

Answer:

x=1/16, 4

Step-by-step explanation:

a=k, b= -(8k+4) , c=81

b²-4ac=0, since real and equal roots

(-(8k+4))² - 4(k)(81)

64k+16+64k-324k=0 (divide all by 4)

64k²-260k+16=0 (factorise). 64= -1/16 and -64/16

therefore x=1/16, 4