Roots of $\frac{{2{P^2}}}{x} + \frac{{3{Q^2}}}{{x - 1}} = 6$ Let $\alpha ,\beta $ are the roots of  $\frac{{2{P^2}}}{x} + \frac{{3{Q^2}}}{{x - 1}} = 6,P,Q \in {R_0}$ where $R_0$ represent non-zero real number
(A) $\alpha ,\beta  \in \left( {0,1} \right)$
(B) $\alpha ,\beta  \in \left( {1,\infty } \right)$
(C) One root lies in $(0, 1)$ and other root lies in $(1, \infty)$
(D) $\alpha ,\beta $ are imaginary roots
My approach is as follow $y = \frac{{2{P^2}}}{x} + \frac{{3{Q^2}}}{{x - 1}} - 6$
Putting $x=\beta$ and $y=0$
$ \frac{{2{P^2}}}{\beta} + \frac{{3{Q^2}}}{{\beta - 1}} = 6$
Putting $x=\alpha$ and $y=0$
$ \frac{{2{P^2}}}{\alpha} + \frac{{3{Q^2}}}{{\alpha - 1}} = 6$
I am not be to proceed as this is not a quaratic equatiom

$2P^2$, $3Q^2$ terms and $6$ in the right hand side are annoying, and the only important thing is that they are positive. So we can let $p = P/\sqrt{3}$ and $q = Q/\sqrt{2}$ to have
$$ \frac{p^2}{x} + \frac{q^2}{x-1} = 1.$$
$x\ne0, 1$, so we have $x(x-1) = p^2(x-1) + q^2 x$, or $$x^2 - (1+p^2+ q^2)x + p^2 =0.$$ Two roots are
$$x_{\pm}=\frac{1+p^2 + q^2 \pm \sqrt{(1+p^2+q^2)^2 - 4p^2 }}{2}. $$
So the situation can be reduced into the problem of quadratic equation.
We can see this in another way. We start with the reduced
$$ f(x) = \frac{p^2}{x} + \frac{q^2}{x-1}.$$
When $x>0$ is small, $q^2/(x-1)$ is just some negative number and $\frac{p^2}{x}$ is very large. (Precisely, we can make this term as big as we want by choose smaller $x>0$.) So at some small $0<x_0<1/2$, $f(x_0)>0$.
When $x<1$ is very close to $1$, $p^2/x$ is just some number and $\frac{q^2}{x-1}$ is negative number with very big absolute value. (like $-10^{100}$.) So at some $1/2<x_1<1$ , $f(x_1)<0$.
Since $f$ is continuous function with $f(x_0)>0>f(x_1)$ with $0<x_0<x_1<1$, there must be a root $y_0\in (x_0, x_1)\subset(0,1)$ with $f(y_0) =0$. This is called the intermediate value theorem. (Google it for further information.)
Similarly by comparing the sign of $f(1+(\text{very small number)})$ and $f(\text{very large number})$, we can assert that there is one other zero $y_1(1, \infty)$.
Since it can be reduced into a quadratic equation $x^2-(1+p^2+q^2)x+p^2$, we can assert that there are 2 or less number of zeros. So $y_0$ and $y_1$ is all of the zeros.
(Thus the answer is option (C).)
