The number of real roots of the equation frac | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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:

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(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

Similar Questions

Solve the given two equations and select the correct option.
$I.$ $42x - 17y = -67$
$II.$ $7x - 12y = -26$

The number of values of $k$ for which the equation $x^2 - 3x + k = 0$ has two real and distinct roots lying in the interval $(0, 1)$ is:

The equation formed by decreasing each root of $ax^2 + bx + c = 0$ by $1$ is $2x^2 + 8x + 2 = 0$. Then:

If $z_1, z_2, z_3, z_4$ are the roots of the equation $z^4 + z^3 + z^2 + z + 1 = 0$,then $\prod_{i=1}^{4} (z_i + 2)$ is equal to:

The equation $\sqrt{x + 1} - \sqrt{x - 1} = \sqrt{4x - 1}$ has

Vedclass Test Series

Exam Paper Generator

Online Exam Module