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