Let $f(x)=\dfrac{3}{x-2}+\dfrac{4}{x-3}+\dfrac{5}{x-4}$, then $f(x)=0$ has roots in the intervals?
I managed to do this by concepts of Location of roots (Quadratic Equation).
But it takes a lot of time also time limit is only 2 min any way to do this in short period of time. Please help.
Two roots only! On $(2,3)$ and on $(3,4)$.
Because our expression is decreasing on all interval from domain and easy to see,
what happens around $2$, around $3$ and around $4$.
Indeed, $f(x)=0$ is equivalent to $ax^2+bx+c=0$, where $a\neq0$.
Thus, this equation has two real roots maximum.
But $\lim\limits_{x\rightarrow2^+}f(x)=+\infty$, $\lim\limits_{x\rightarrow3^-}f(x)=-\infty$ and $f$ is a decreasing continuous function on $(2,3)$.
Thus, our equation has an unique root on $(2,3)$.
By the same way we'll get that our equation has an unique root on $(3,4)$ and we are done!