Let а, b, с the be the lengths of sides of a triangle(no two of them equal) and k ∈ R.If the roots of the equation x2 + 2(a + b + с)х + 6k(ab + bc + ca) = 0 real, then:
1
k > 1
2
k < \(\frac{1}{4}\)
3
k > \(\frac{2}{3}\)
4
k < \(\frac{2}{3}\)
5
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