The number of real roots of the equation frac | vedclass

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(B) Given the equation: $\frac{P^2}{x} + \frac{Q^2}{x - 1} = 1$.Multiplying by $x(x - 1)$,we get: $P^2(x - 1) + Q^2x = x(x - 1)$.Expanding the terms:

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$2$

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(B) Given the equation: $\frac{P^2}{x} + \frac{Q^2}{x - 1} = 1$.Multiplying by $x(x - 1)$,we get: $P^2(x - 1) + Q^2x = x(x - 1)$.Expanding the terms: $P^2x - P^2 + Q^2x = x^2 - x$.Rearranging into a standard quadratic form $ax^2 + bx + c = 0$: $x^2 - (P^2 + Q^2 + 1)x + P^2 = 0$.The discriminant $D$ of this quadratic equation is given by $D = b^2 - 4ac$.Here,$a = 1$,$b = -(P^2 + Q^2 + 1)$,and $c = P^2$.$D = (P^2 + Q^2 + 1)^2 - 4(1)(P^2) = (P^2 + Q^2 + 1)^2 - 4P^2$.Using the identity $a^2 - b^2 = (a - b)(a + b)$,we get $D = (P^2 + Q^2 + 1 - 2P)(P^2 + Q^2 + 1 + 2P) = ((P - 1)^2 + Q^2)((P + 1)^2 + Q^2)$.Since $P$ and $Q$ are non-zero real numbers,$D > 0$. Therefore,the equation has two distinct real roots.

The number of real roots of the equation $\frac{P^2}{x} + \frac{Q^2}{x - 1} = 1$,where $P$ and $Q$ are non-zero real numbers,is

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