Given equation e4x+4e3x−58e2x+4ex+1=0
Let f(x)=e2x(e2x+e2x1+4(ex+ex1)−58)
Now let ex+ex1=t
i.e. t2+4t−58=0
t=2−4±16+4×58
=2−4±262
t=−2+262 is possible and t=−2−262 is not possible as t≥2
i.e. ex+ex1=−2+262
e2x−(−2+262)ex+1=0
For real solution, discriminant 248−862>0
Also 2a−b=2−2+262=62−1>0
Hence, both roots of the equation are positive real roots.