What are the solutions of $frac{pi^e}{x-e}+frac{e^pi}{x-pi}+frac{pi^pi+e^e}{x-pi-e}=0$?

Question

Prove that $$frac{pi^e}{x-e}+frac{e^pi}{x-pi}+frac{pi^pi+e^e}{x-pi-e}=0$$ has one real root in$(e,pi)$ and other in $(pi,pi+e)$.

For $xin (-infty,e)$ the equation will be always be negative and for $xin (pi+e,infty)$ the equation will be always positive so roots must lie in $(e,pi+e)$.
Now how to do for specific intervals?
Like, if we could prove equation is negative for x=$e^+$ and positive for $x=pi^-$,then one real root would lie in this interval.
But how do we do this?

Z Ahmed · Answer

The roots of $$f(x)=frac{pi^e}{x-e}+frac{e^pi}{x-pi}+frac{pi^pi+e^e}{x-pi-e}=0$$
and $$g(x)=pi^e(x-pi)(x-pi-e)+e^{pi}(x-e)(x-e-pi)+(pi^pi+e^e)(x-e)(x-pi)=0$$
would coincide
Note that $$g(e)=pi^{1+e}(pi-e)>0,~~ g(pi)=e^{pi+1}(pi-e)<0,~~ g(e+pi)=epi(pi^pi+e^e)>0$$ This proves that the quadratic $g(x)$ has one real root in $(e,pi)$ and other one in $(pi, pi+)e.$ So will be the case for $f(x)=0.$

What are the solutions of $frac{pi^e}{x-e}+frac{e^pi}{x-pi}+frac{pi^pi+e^e}{x-pi-e}=0$?

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