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

Mathematics Asked on January 1, 2022

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?

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 $$f(x)=0.$$

Answered by Z Ahmed on January 1, 2022

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