Roots of $\frac{{2{P^2}}}{x} + \frac{{3{Q^2}}}{{x - 1}} = 6$

Question

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

Answer 1

$2P^2$, $3Q^2$ 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).)

Answer 2

EDIT (1/18) -- My original argument here had a serious oversight for certain possibilities for values of the roots (the hazard of "wee-hours mathematics"...) that is not easily "fixed". So I withdraw my earlier answer and will take another approach.

Consider the solutions of the equation $ \ \frac{{2{P^2}}}{x} + \frac{{3{Q^2}}}{{x - 1}} \ = \ 6 \ \ $ as the intersections of two "rectangular" hyperbolas, as given by $ \ \frac{{2{P^2}}}{x} \ = \ 6 - \frac{{3{Q^2}}}{{x - 1}} \ \ . $ The former hyperbola has the horizontal asymptote $ \ y = 0 \ $ and the vertical asymptote $ \ x = 0 \ \ ; $ the function $ \ f(x) \ $ that it represents is negative only for $ \ x < 0 \ $ and positive only for $ \ x > 0 \ \ . $ The latter is "vertically stretched" relative to the former, since it has a different positive coefficient. The significant properties for our argument are that it is also translated "horizontally" and "vertically" in the plane and also "flipped" about its horizontal asymptote. Consequently, it has a horizontal asymptote of $ \ y = 6 \ , \ $ a vertical asymptote of $ \ x = 1 \ , \ $ and its corresponding function $ \ g(x) \ $ is greater than $ \ 6 \ $ for $ \ x < 1 \ $ and less than $ \ 6 \ $ for $ \ x > 1 \ \ . $ Each function is continuous over its domain.

Plainly, the two function curves have no intersections for $ \ x < 0 \ \ , $ as $ \ f(x) < 0 \ , \ g(x) > 6 \ \ . $ In the interval $ \ 0 < x < 1 \ \ , \ f(x) $ is decreasing from "positive infinity", while $ \ g(x) \ $ is increasing from $ \ 6 \ $ toward "positive infinity" as it approaches its vertical asymptote "from the left". So there is a single intersection of the function curves in this interval.

For $ \ x > 1 \ \ , \ f(x) \ $ continues its approach "from above" to the $ \ x-$axis, while $ \ g(x) \ $ is increasing from "negative infinity" to its asymptotic value of $ \ 6 \ \ . $ The two function curves will again cross once only somewhere "to the right" of the vertical asymptote of $ \ g(x) \ \ . $ (If $ \ |P| < 1 \ \ , $ this occurs close to the $ \ x-$axis; for $ \ |P| >> 1 \ \ , $ the intersection is close to $ \ y = 6 \ \ . ) $ We can be assured that there is only one intersection in each of these intervals since $ \ f'(x) \ = \ -\frac{2P^2}{x^2} \ < \ 0 \ \ , $ so $ \ f(x) \ $ is always decreasing over its domain ("wrapping around infinity" in the neighborhood of $ \ x = 0 \ ) \ , $ while $ \ g'(x) \ = \ \frac{3Q^2}{(x-1)^2} \ > \ 0 \ \ , $ hence $ \ g(x) \ $ is always increasing with a "wrap-around " in the neighborhood of $ \ x = 1 \ ) \ . \ $

Thus the original equation has one root which is greater than $ \ 1 \ $ and the other is between $ \ 0 \ $ and $ \ 1 \ \ $ [ choice $ \ \mathbf{(C)} \ ] \ \ . $

$$ \ \ $$

If we multiply the curve equation through by $ \ x·(x-1) \ \ , $ we obtain the quadratic equation $$ 2P^2·(x - 1) \ + \ 3Q^2·x \ \ = \ \ 6·x·(x-1) \ \ \rightarrow \ \ 6x^2 \ - \ [6 + 2P^2 + 3Q^2]·x \ + \ 2P^2 \ \ = \ \ 0 \ \ . $$ The discriminant is not particularly pleasant to work with (other approaches would be preferable for a "timed exam"), but can be made to provide some insight: $$ \ \Delta \ \ = \ \ (36 + 4P^4 + 9Q^4 + 24P^2 + 36Q^2 + 12P^2Q^2) \ - \ 48P^2 \ \ . $$ This can be re-arranged in the expression for the square of the difference between the roots as $$ (\alpha \ - \ \beta)^2 \ \ = \ \ \frac{\Delta}{6^2} \ \ = \ \ \frac{P^2·(4P^2 - 24) \ + \ Q^2·(12P^2 + 9Q^2 + 36) \ + \ 36}{36} \ \ . $$ It is not immediately obvious that the difference between the roots must always be greater than $ \ 1 \ \ , $ since the first term can be negative. If we graph the expression in the numerator in the $ \ PQ-$plane as the curve equation $ \ P^2·(4P^2 - 24) \ + \ Q^2·(12P^2 + 9Q^2 + 36) \ = \ c \ \ , $ we obtain a curve resembling a lemniscate -- a bi-lobed curve centered at the origin with its "major axis" on the $ \ x-$axis for $ \ c = 0 \ \ . $ As we decrease $ \ c \ \ , $ the two lobes separate and shrink, "degenerating" to the points $ \ P = \pm \sqrt3 \ , \ Q \ = \ 0 \ $ for $ \ c \ = \ -36 \ \ . $ At this minimum value then, $ \ \Delta \ = \ 0 \ \ $ and we would have $ \ \alpha \ , \ \beta \ = \ \frac{6 \ + \ 2·3}{2·6} \ = \ 1 \ \ . $

However, it is specified that $ \ P \ , \ Q \ \neq \ 0 \ \ , $ so we necessarily have $ \ \Delta > 0 \ \ . $ Hence the original curve equation have two roots. With $ \ P \ = \ \sqrt3 \ \ $ and $ \ Q \ = \ \pm \varepsilon \ \ , $ the two roots "separate" into $$ \alpha \ , \ \beta \ \ = \ \ \frac{12 + 3·\varepsilon^2}{12} \ \pm \ \frac{\sqrt{3·(-12) \ + \ \varepsilon^2·(36 + 9·\varepsilon^2 + 36) \ + \ 36}}{12} $$ $$ \approx \ \ 1 \ \pm \ \frac{6·\sqrt{2}·\varepsilon}{12} \ \ = \ \ 1 \ \pm \ \frac{ \sqrt{2}}{2}\varepsilon \ \ , $$ which places a root of the original curve equation "on either side" of $ \ x \ = \ 1 \ \ . $ (Further exploration indicates that this holds for all permitted values of $ \ P \ $ and $ \ Q \ \ , $ but I won't pursue that here.)