If b > a, then the equation (x - a) (x - b) - 1 = 0 has

If $$\alpha \,\text{and}\,\beta $$ $$(\alpha \, < \,\beta )$$ are the roots of the equation $${x^2} + bx + c = 0\,$$, where $$c < 0 < b$$, then

For the equation $$3{x^2} + px + 3 = 0$$. p > 0, if one of the root is square of the other, then p is equal to

For $$2 \le r \le n,\,\,\,\,\left( {\matrix{ n \cr r \cr } } \right) + 2\left( {\matrix{ n \cr {r - 1} \cr } } \right) + \left( {\matrix{ n \cr {r - 2} \cr } } \right) = $$