Prove the following for the given expression Consider the equation $$\frac{\pi ^e}{(x-e)} + \frac{e^{\pi}}{(x-\pi)} + \frac{\pi^{\pi}+e^e}{(x-e-\pi)}=0$$
Prove that 
1) This equation has two real roots in $(\pi-e, \pi +e) $
2) This equation has one real root in $(e,\pi)$ and other root in $(\pi, e+\pi)$
My approach : Let $$f(x)=\frac{\pi ^e}{(x-e)} + \frac{e^{\pi}}{(x-\pi)} + \frac{\pi^{\pi}+e^e}{(x-e-\pi)}$$. Then I tried to find the sign of $f(\pi-e)f(\pi +e)$ so as to check for its sign to prove the first part. But this gives me a very tedious equation from which determining the sign is difficult. Moreover $f(\pi +e)$ creates a 0 in the denominator in third term.Any help would be greatly appreciated.
Edit 1: Can anyone prove this using the properties and fundamentals just of polynomials. I know that maths is meant to learn what you don't understand but at current I want to know if this question can be solved using only the fundamentals of algebra and the polynomials without use of calculus
 A: multiply out the term
$$\pi^e(x-\pi)(x-e-\pi)+e^\pi(x-e)(x-e-\pi)+(\pi^\pi+e^e)(x-e)(x-\pi)=...$$
ok then consider
$$f(x)=\frac{\pi^e}{x-e}+\frac{e^\pi}{x-\pi}+\frac{\pi^\pi+e^e}{x-e-\pi}$$
and compute $$f(\pi-e)\approx -27.7935443004972883808036072099123560713149915<0$$ and $$\lim_{x \to \pi+e}f(x)=+\infty$$
A: Multiplying out the denominator into the numerator we get:
$$f(x)=\pi^e(x-\pi)(x-e-\pi)+e^\pi(x-e)(x-e-\pi)+(\pi^\pi+e^e)(x-e)(x-\pi)$$
I'll show you how to prove the part about $(e,\pi)$, the other proofs are similar. This proof does not use any complex quadratic formula or tough arithmetic calculation.
Take $x_1=e+h$, where $h$ is a very small number $\approx10^{-20}>0$. Compute 
$$f(x_1)=\color{green}{\pi^e(e-\pi+h)(-\pi+h)}+\color{blue}{e^\pi(h)(-\pi+h)}+\color{blue}{(\pi^\pi+e^e)(h)(e-\pi+h)}$$
Since $h$ is very very small, we can almost equate $\color{blue}{h\cdot\text{constant value}\approx0}$ and $\color{green}{\text{constant}+h\approx\text{constant}}$. Hence, we get:
$$f(x_1)\approx\pi^e(e-\pi)(-\pi)>0$$
Now, take $x_2=\pi-h$. Computing $f(x_2)$ by similar logic, we get:
$$f(x_2)=e^\pi(\pi-e)(-e)<0$$
Since $f(x)$ has changed sign in the interval $(e,\pi)$, this implies that it must have crossed the x-axis at some $x_0\in(e,\pi)$ as it is a continuous function. Hence, $f(x_0)=0$ and thus $f(x)$ has a root in $(e,\pi)$.

For the first proof, you'll need to break the given interval of $(\pi-e,\pi+e)$ at some suitable value of $x'$. Then use the same logic as described above, to prove the existence of one real root each in $(\pi-e,x')$ and $(x',\pi+e)$, for a total of two real roots. 

Can you solve this on your own from here now?

PS: (since you mentioned you don't have much experience of limits) The technique of taking $h$ is essentially me showing you how to take the Left hand Limits and Right Hand Limits of $f(x)$ without actually mentioning the scary terminology! ;)
A: Let $x\notin\{e,\pi,e+\pi\}$ then multiplying both sides by $(x-e)(x-\pi)(x-e-\pi)$ gives
$$(x-e)(x-\pi)(x-e-\pi)f(x)=g(x)$$
where 
$$g(x):=\pi^e(x-\pi)(x-e-\pi)+e^\pi(x-e)(x-e-\pi)+(\pi^\pi+e^e)(x-e)(x-\pi)$$
Clearly $g(x)$ is a polynomial of degree two hence a quadratic polynomial. Moreover the coefficients are all real so if one root of $g(x)$ is real the other one must be real as well. Now you can do a sign check of $g(x)$. Since we excluded $e,\pi,e+\pi$ we are only allowed to approach these values in our argument. So if $x$ is close to $e$ from the right side the second and third term are negligible since $x-e$ becomes smaller and smaller almost zero. So the first term dominates. Since $e<\pi$ then $g(x)>0$ for $x$ close to $e$. Similarly we approach $\pi$ from the left. Again the first and the third term become negligibly small so the middle term dominates. Since $\pi>e$ then $g(x)<0$ for $x$ close to $\pi$. But $g(x)$ is a polynomial so a continuous function therefore there exists some $x_1\in(e,\pi)$ where $g(x_1)=0$ (by Intermediate Value Theorem). The same analysis for the interval $(\pi,e+\pi)$. These are the only two roots of $g(x)$. But from the main equation that relates $f(x)$ and $g(x)$ it then follows that indeed these are the roots also for $f(x)$.
A: Let $f(x)=\frac{\pi ^e}{(x-e)} + \frac{e^{\pi}}{(x-\pi)} + \frac{\pi^{\pi}+e^e}{(x-e-\pi)}.$
Thus, $f$ is a continuous function on $(\pi,\pi+e)$ and on $(e,\pi)$ and calculate four limits of $f$ in these bounds. 
For example, $$\lim_{x\rightarrow\pi+e^-}f(x)=-\infty$$ and
$$\lim_{x\rightarrow\pi^+}f(x)=+\infty$$
A: First, we multiply the expression by each of the denominators to rewrite the expression as $\pi^e(x-\pi)(x-e-\pi)+e^\pi(x-e)(x-e-\pi) + (\pi^{\pi}+e^e)(x-e)(x-\pi)=0$ where $x$ is not $e, \pi$, or $e + \pi$ (since the original expression is not defined for those values). Call that polynomial $g(x)$. When this polynomial has a root that is not $e, \pi$, or $e + \pi$, then such a root is a solution to the original equation. So lets use this to answer the questions (note that $\pi > e > \pi - e$):
1) If we have part 2 then part 1 follows since if we have a solution in $(e,\pi)$ and another solution in $(\pi,e+\pi)$, then the two solutions are both contained in $(e,e+\pi)$ which is a subset of $(\pi-e,\pi+e)$ since $e>\pi-e$.
2) Notice that:


*

*$g(e) = \pi^e(e-\pi)(-\pi)>0$

*$g(\pi) = e^\pi(\pi - e)(-e)<0$

*$g(e + \pi) = (\pi^{\pi}+e^e)(\pi)(e)>0$


So $g(x)$ must be $0$ for some $x$ in $(e,\pi)$ and for another $x$ in $(\pi,e+\pi)$. Note that neither of these solutions are any of $e, \pi$, or $e + \pi$ as required.
A: For $x\neq b,d,f$, we have the equation $$\begin{align}\frac a{x-b}+\frac c{x-d}+\frac g{x-f}=0&\implies \small a(x-d)(x-f)+c(x-b)(x-f)+g(x-b)(x-d)=0\\&\implies\small(a+c+g)x^2-[a(d+f)+c(b+f)+g(b+d)]x+adf+bcf+bdg=0\end{align}$$ Hence

$$\begin{align}\small x&=\small\frac{[a(d+f)+c(b+f)+g(b+d)]\pm\sqrt{[a(d+f)+c(b+f)+g(b+d)]^2-4(a+c+g)(adf+bcf+bdg)}}{2(a+c+g)}\end{align}$$

Substituting $a=\pi^e$, $b=e$, $c=e^\pi$, $d=\pi$, $g=\pi^\pi+e^e$ and $f=e+\pi$, we get $$x=\frac{703.135601...\pm\sqrt{494399.674629...-475524.453356...}}{194.432544...}=2.909741...,4.322952...$$

Now $(\pi-e,e+\pi)=(0.42...,5.85...)$ and we see that both solutions lie in this interval.
Also, $2.909741...\in(e,\pi)$ and $4.322952\in(\pi,e+\pi)$ and so the proof is complete.

