If x, y and z are real and different and $$\,u = {x^2} + 4{y^2} + 9{z^2} - 6yz - 3zx - 2xy$$, then u is always.

Let a > 0, b > 0 and c > 0. Then the roots of the equation $$a{x^2} + bx + c = 0$$

If $$\ell $$, m, n are real, $$\ell \ne m$$, then the roots by the equation :
$$(\ell - m)\,{x^2} - 5\,(\ell + m)\,x - 2\,(\ell - m) = 0$$ are

Given that $${C_1} + 2{C_2}x + 3{C_3}{x^2} + ......... + 2n{C_{2n}}{x^{2n - 1}} = 2n{\left( {1 + x} \right)^{2n - 1}}$$
where $${C_r} = {{\left( {2n} \right)\,!} \over {r!\left( {2n - r} \right)!}}\,\,\,\,\,r = 0,1,2,\,............,2n$$
Prove that $${C_1}^2 - 2{C_2}^2 + 3{C_3}^2 - ............ - 2n{C_{2n}}^2 = {\left( { - 1} \right)^n}n{C_n}.$$