Let P(x) = 0 be a 50th degree polynomial with 50 real and distinct roots say  ${\alpha }_1,{\alpha }_2,{\alpha }_3,\mathrm{......},{\alpha }_{50}$. Then  $P\left(x\right)=A(x-{\alpha }_1)(x-{\alpha }_2)(x-{\alpha }_3)\mathrm{.....}(x-{\alpha }_{50})$, where  $A\in R-\left\{0\right\}$ and  ${\alpha }_i\ne 0\forall i\in [1,50]$.The roots of the equation  $50p\left(x\right)p^{\left|\right|}\left(x\right)=49{\left(p{}^{|}\left(x\right)\right)}^2$ are

$$\begin{gathered} 50p\left(x\right)p^{\left|\right|}\left(x\right)=(50-1){\left(p{}^{|}\left(x\right)\right)}^2 \\ \Rightarrow 50[p\left(x\right)p^{\left|\right|}\left(x\right)-{\left(p^{|}\left(x\right)\right)}^2]={\left(p{}^{|}\left(x\right)\right)}^2 \\ \Rightarrow 50\frac{p\left(x\right)p^{\left|\right|}\left(x\right)-{\left(p^{|}\left(x\right)\right)}^2}{{\left(p\left(x\right)\right)}^2}+{\left(\frac{p^{|}\left(x\right)}{p\left(x\right)}\right)}^2=0 \\ \Rightarrow 50d\left(\frac{p^{|}\left(x\right)}{p\left(x\right)}\right)+{\left(\frac{p^{|}\left(x\right)}{p\left(x\right)}\right)}^2=0 \dots \dots .. \left(1\right) \\ \mathcal{l}np\left(x\right)=\mathcal{l}nA+\mathcal{l}n(x-{\alpha }_1) \dots \dots + \mathcal{l}n(x-{\alpha }_{50}) \\ \frac{p^{|}\left(x\right)}{p\left(x\right)}=\frac{1}{x-{\alpha }_1}+\frac{1}{x-{\alpha }_2}+\mathrm{.......}+\frac{1}{x-{\alpha }_{50}}=\sum _{i=1}^{50}\frac{1}{x-{\alpha }_i} \\ \frac{p^{|}\left(x\right)}{p\left(x\right)}=\sum _{i=1}^{50}\frac{1}{(x-{\alpha }_i)} \dots \dots .. \left(2\right) \\ {\left(\frac{p^{|}\left(x\right)}{p\left(x\right)}\right)}^{|}=-\sum _{i=1}^{50}\frac{1}{{(x-{\alpha }_i)}^2}\dots \dots .. \left(3\right) \end{gathered}$$

Substitute (2) and (3) in equation (1) 

$50\sum _{i=1}^{50}\frac{1}{{(x-{\alpha }_i)}^2}={\left(\sum _{i=1}^{50}\frac{1}{x-{\alpha }_i}\right)}^2$

If all roots are equal then LHS = RHS but all roots are distinct, so roots are imaginary.