MHB No. of real roots of Quadratic equation

AI Thread Summary
The equation given is analyzed by defining the function f(x) as the sum of three decreasing functions, each with points of discontinuity. In the interval (e, π), f(x) decreases from +∞ to -∞, indicating the presence of exactly one real root. Similarly, in the interval (π, π+e), f(x) also decreases, suggesting another real root exists there. The overall conclusion is that the equation has two real roots, one in each specified interval. Understanding the behavior of f(x) is crucial for determining the number of real roots.
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The no. of Real Roots of the equation $\displaystyle \frac{\pi^e}{x-e}+\frac{e^\pi}{x-\pi}+\frac{\pi^{\pi}+e^{e}}{x-\pi-e} = 0$

My try:: Let $\displaystyle f(x) = \frac{\pi^e}{x-e}+\frac{e^\pi}{x-\pi}+\frac{\pi^{\pi}+e^{e}}{x-\pi-e}$

Now we will take a interval $x\in \left(e\;,\pi\right)$ and $x\in \left(\pi\;,\pi+e\right)$

$\bullet$ If $x\in (e,\pi),$ Then $(x-e) > 0$ and $(x-\pi) < 0$

So $f(x) = $

Similarly

$\bullet$ If $x\in (\pi,\pi+e),$ Then $(x-\pi) > 0$ and $(x-\pi-e) < 0$

So $f(x) = $

y Question is How Can I check Sign of $f(x)$ in Given Interval and Is Iam Thinking Right.

If Not please explain me,

Thanks
 
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Re: no. of real roots of Quadratic equation

Have you considered looking at the discriminant?
 
Re: no. of real roots of Quadratic equation

jacks said:
The no. of Real Roots of the equation $\displaystyle \frac{\pi^e}{x-e}+\frac{e^\pi}{x-\pi}+\frac{\pi^{\pi}+e^{e}}{x-\pi-e} = 0$

My try:: Let $\displaystyle f(x) = \frac{\pi^e}{x-e}+\frac{e^\pi}{x-\pi}+\frac{\pi^{\pi}+e^{e}}{x-\pi-e}$

Now we will take a interval $x\in \left(e\;,\pi\right)$ and $x\in \left(\pi\;,\pi+e\right)$

$\bullet$ If $x\in (e,\pi),$ Then $(x-e) > 0$ and $(x-\pi) < 0$

So $f(x) = $

Similarly

$\bullet$ If $x\in (\pi,\pi+e),$ Then $(x-\pi) > 0$ and $(x-\pi-e) < 0$

So $f(x) = $

y Question is How Can I check Sign of $f(x)$ in Given Interval and Is Iam Thinking Right.

If Not please explain me,

Thanks
It looks to me as though you are thinking along the right lines here. Each of the three functions $\dfrac{\pi^e}{x-e}$, $\dfrac{e^\pi}{x-\pi}$, $\dfrac{\pi^{\pi}+e^{e}}{x-\pi-e}$ is a decreasing function, except at the points where it has a discontinuity. For example, the first of those functions, $\dfrac{\pi^e}{x-e}$, decreases from $0$ to $-\infty$ in the interval $(-\infty, e)$, and then decreases from $+\infty$ to $0$ in the interval $(e,\infty).$

When you add the three functions together, their sum $f(x)$ will also be a decreasing function everywhere except at its points of discontinuity. So for example it will decrease from $+\infty$ to $-\infty$ in the interval $(e,\pi)$, and therefore it must have exactly one zero in that interval.
 
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