Question

Hii guys!!....solve this question...​

Answer 1

hope it may help you thank you have a NYC day

Answer 2

$let \: p(x) = kx^{2} + 2x + 1 = 0$
Here , a = k , b = 2 and c = 1.
Since,P(x) has real and distint roots.
Therefore,
D > 0
$b ^{2} - 4ac > 0$
$2 ^{2} - 4 \times k \times 1 > 0$
$4 - 4k > 0$
$4 < 4k$
$1 < k$
or
$k > 1$