$If P is a cons\tan t$,  $f\left(x\right)=\left|\begin{array}{ccc}{x}^2& x^3& x^4\\ 2& 3& 6\\ P& P^2& P^3\end{array}\right|$ $and$$f"\left(x\right)=0$   $has roots$ $\alpha ,\beta$ ,$then$

$\begin{array}{l}f"\left(x\right)=\left|\begin{array}{ccc}2& 6x& 12x^2\\ 2& 3& 6\\ p& p^2& p^3\end{array}\right|=0\\ Expand the determinant\\ \Rightarrow 12\left(2p^2-3p\right)x^2-6x\left(2p^3-6p\right)+6p^2\left(p-2\right)=0\\ \Rightarrow 12p\left(2p-3\right)x^2-12p\left(p^2-3\right)+6p^2\left(p-2\right)=0\\ \text{At }P=0,\text{ its an identity}\\ \text{At }P=\sqrt{3},\frac{b}{a}=0,\text{so roots are opposite in sign and equal in magnitude}\\ \text{At }P=2,\frac{c}{a}=0,\text{ so product of the roots is zero. }\\ \text{At }P=-\sqrt{3},\frac{c}{a}=\frac{6p^2\left(p-2\right)}{12p\left(2p-3\right)}\text{\hspace{0.17em}negative}\\ \text{So product of the roots is negative}\end{array}$