If one root of the equation $x^2-30x+p=0$  square of the other then p is equal to

Let  ${\alpha }^2$  and $\alpha$ be the roots of the equation

 $x^2-30x+p=0 ... ( i )$

 Then,   ${\alpha }^4-30{\alpha }^2+\rho =0 ... ( ii )$

and  ${\alpha }^2-30\alpha +p=0$      ... (iii)  

  On eliminating p from Eqs. (ii) and (iii), we get  

 $\begin{array}{lcl} & & {\alpha }^4-30{\alpha }^2-\left({\alpha }^2-30\alpha \right)=0\\ \Rightarrow & {\alpha }^4-30{\alpha }^2-{\alpha }^2+30\alpha =0& \\ \Rightarrow & \alpha \left[\left({\alpha }^3-30\alpha -\alpha +30\right)\right]=0& \\ \Rightarrow & \left.\alpha \left[{\alpha }^3-31\alpha +30\right)\right]=0& \\ \Rightarrow & \alpha (\alpha -1)\left({\alpha }^2+\alpha -30\right)=0& \\ \Rightarrow & \alpha (\alpha -1)(\alpha -5)(\alpha +6)=0& \\ \Rightarrow & \alpha =0,1,5,-6& \\ \Rightarrow & \alpha =5\text{ or }-6[\because \alpha =0,1\text{ not possible }]& \end{array}$

 From Eq. (i), product of the roots $={\alpha }^2\cdot \alpha =\rho$