Question

Find value of K,kx2+2x+1=0 has real and distinct roots

Answer 1

Answer :

k<1

Explanation:

Compare kx²+2x+1=0 with

ax²+bx+c=0 , we get

a=k , b = 2 , c= 1

Now,

Discreminant (D)>0

[ Equal roots, given ]

=> b² - 4ac > 0

=> 2²-4×k×1>0

=> 4 - 4k>0

=> -4k > -4

=> k < 4/4

=> k <1

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