Question

find
the value of k for which the
given
equation has real
roots
3x^2+2k+1=0

​

Answer 1

Answer:

$3 {x}^{2} + 2k + 1 = 0 \\ \\ g(x) = 0 \\ x + 1 = 0 \\ x = - 1 \\ \\ 3( - 1 {)}^{2} + 2k + 1 = 0 \\ 2k = - 1 - 3 \\ therefore \: \: = - 4 \div 2 \\ \\ answer \: is \: - 2$

Answer 2

b2-4ac > 0(condition for real root

2k^2-4×3×1>0

4k^2-12>0

4k^2>12

K^2>3

K>√3

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