  Let -1\leq B<A\leq 1. Condition on \beta, is determined so that 1+\beta
zp'(z)/p^k(z)\prec(1+Az)/(1+Bz)\;(-1<k\leq3) implies p(z)\prec \sqrt{1+z}.
Similarly, condition on \beta is determined so that 1+\beta zp'(z)/p^n(z) or
p(z)+\beta zp'(z)/p^n(z)\prec\sqrt{1+z}\;(n=0, 1, 2) implies
p(z)\prec(1+Az)/(1+Bz) or \sqrt{1+z}. In addition to that condition on \beta is
derived so that p(z)\prec(1+Az)/(1+Bz) when p(z)+\beta
zp'(z)/p(z)\prec\sqrt{1+z}. Few more problems of the similar flavor are also
considered.
