For what value of k does the equation x2+2x+k2+1=0 has a real and equal root
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Answered by
44
if equation have real and equal roots
then Discriminant of equation must be equal to zero
e.g D = b² - 4ac = 0
now, here equation, x² +2x +k² +1 =0
b = 2
a = 1
c = k² +1
use this in above
D = (2)² -4(1)(k² +1) =0
4 -4k² -4 =0
K ² =0
K =0
hence value of K =0
then Discriminant of equation must be equal to zero
e.g D = b² - 4ac = 0
now, here equation, x² +2x +k² +1 =0
b = 2
a = 1
c = k² +1
use this in above
D = (2)² -4(1)(k² +1) =0
4 -4k² -4 =0
K ² =0
K =0
hence value of K =0
abhi178:
first you refresh answer!!!
in last step (2)² -4(k²+1) =0
4 -4k² -4 =0
(4 -4) -k² =0
k² =0
k =0
k=0
Sorry but he did u got 4-k*-1
Sorry hw
some mostake occur by me , now this is correct
Answered by
10
x^2+2x+k^2+1=0
as it has real and equal roots discriminat is 0
b^2-4ac=0
2^2-4(1)(k^2+1)=0
4-4k^2-4=0
-4k^2=o
k^2=o
k=0
as it has real and equal roots discriminat is 0
b^2-4ac=0
2^2-4(1)(k^2+1)=0
4-4k^2-4=0
-4k^2=o
k^2=o
k=0
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