# No. of real roots of Quadratic equation

#### jacks

##### Well-known member
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 plz explain me,

Thanks

Last edited:

#### MarkFL

Staff member
Re: no. of real roots of Quadratic equation

Have you considered looking at the discriminant?

#### Opalg

##### MHB Oldtimer
Staff member
Re: no. of real roots of Quadratic equation

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 plz 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.