Question

Find K...... ......help to solve

Answer 1

Given,
Kx2+1-2(k-1)x+x2
=x2(k+1)-2(k-1)+1
 
Here,
a=(k+1)
b=-2(k-1) & c=1
for real and equal roots,
d=0=b2-4ac=0
putting the values of a,b,c
[-2(k-1)]2-4(k+1)(1)=0
Then on opening the brackets we’ll get,
(4k)(k-3)=0
= 4k=0  and  (k-3)=0
= k=0 or k=3


ℍᝪℙℰ ⅈᝨ ℍℰℒℙՏ
 ℽᝪႮ 

Answer 2

Answer:

ok

0,3

ok

thank you

ans

ok

bukja