x^2+kx+k=0 has no solution then the value of k
rajkumaryaduvanshi21:
full solution of my question
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Answered by
1
k is equal to zero and is not
greater than zero
greater than zero
sir full solve
sry i can only tell answers as it takes time
and my answer is wrong
Answered by
16
x^2+kx+k=0
this equation has two distinct real solution.
now,
= b^2 -4ac = 0
if b = 1 , a = 1.
= k^2 - 4*(1)*k > 0
= k^2 -4k. > 0
= k (k-4) > 0
then ,
k >0 and k> 4
this equation has two distinct real solution.
now,
= b^2 -4ac = 0
if b = 1 , a = 1.
= k^2 - 4*(1)*k > 0
= k^2 -4k. > 0
= k (k-4) > 0
then ,
k >0 and k> 4
answer is right
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