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: $\frac{P^2}{x} + \frac{Q^2}{x - 1} = 1$Multiply by $x(x - 1)$ to clear the denominators:$P^2(x - 1) + Q^2(x) = x(x - 1)$$P^2x

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(B) Given the equation: $\frac{P^2}{x} + \frac{Q^2}{x - 1} = 1$Multiply by $x(x - 1)$ to clear the denominators:$P^2(x - 1) + Q^2(x) = x(x - 1)$$P^2x - P^2 + Q^2x = x^2 - x$Rearranging into a standard quadratic form $ax^2 + bx + c = 0$:$x^2 - (P^2 + Q^2 + 1)x + P^2 = 0$The discriminant $D$ of this quadratic equation is given by $D = b^2 - 4ac$:$D = (P^2 + Q^2 + 1)^2 - 4(1)(P^2)$$D = (P^2 + Q^2 + 1)^2 - 4P^2$Using the identity $a^2 - b^2 = (a - b)(a + b)$:$D = (P^2 + Q^2 + 1 - 2P)(P^2 + Q^2 + 1 + 2P)$$D = ((P - 1)^2 + Q^2)((P + 1)^2 + Q^2)$Since $P$ and $Q$ are non-zero real numbers,$Q^2 > 0$. Thus,$D > 0$ for all non-zero real $P$ and $Q$.Since the discriminant is strictly positive,the quadratic equation has $2$ 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

What is the minimum value of the expression $x^2 + 4y^2 + 3z^2 - 2x - 12y - 6z + 14$?

If the roots of the equation $\frac{1}{x + p} + \frac{1}{x + q} = \frac{1}{r}$ are equal in magnitude but opposite in sign,then the product of the roots will be

Let $\alpha, \beta$ be the roots of the equation $x^{2}-\sqrt{2}x+\sqrt{6}=0$ and $\frac{1}{\alpha^{2}}+1, \frac{1}{\beta^{2}}+1$ be the roots of the equation $x^{2}+ax+b=0$. Then the roots of the equation $x^{2}-(a+b-2)x+(a+b+2)=0$ are...

If $p, q, r$ are in $A.P.$ and are positive,the roots of the quadratic equation $px^2 + qx + r = 0$ are all real for

If $a, b$ and $c$ are in arithmetic progression,then the roots of the equation $a x^{2}-2 b x+c=0$ are

Vedclass Test Series

Exam Paper Generator

Online Exam Module