find the values of k for which the given equation has real and equal root. x^2(k + 1) - 2(k - 1) + 1 = 0
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5
Answer:
if equal roots
then b^2-4ac=0
4(k-1)^2-4(k+1)(1)=0
(k-1)^2=k+1
k^2-2k+1=k+1
k^2-3k=0
k(k-3)=0
k=0,3
rahman786khalilu:
ty for brainliest
Answered by
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Answer:-
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