The solution set of the equation $pqx^2 - (p + q)^2x + (p + q)^2 = 0$ is

  • A
    $\left\{ \frac{p}{q}, \frac{q}{p} \right\}$
  • B
    $\left\{ pq, \frac{p}{q} \right\}$
  • C
    $\left\{ \frac{q}{p}, pq \right\}$
  • D
    $\left\{ \frac{p + q}{p}, \frac{p + q}{q} \right\}$

Explore More

Similar Questions

What is the number of real solutions to the equation $(5 + 2\sqrt{6})^{x^2 - 3} + (5 - 2\sqrt{6})^{x^2 - 3} = 10$?

Let $f(x)$ be a polynomial and $a, b$ be distinct real numbers. Then the remainder in the division of $f(x)$ by $(x-a)(x-b)$ is

Let $\alpha$ and $\beta$ be the roots of the quadratic equation $x^2 \sin \theta - x(\sin \theta \cos \theta + 1) + \cos \theta = 0$ where $0 < \theta < 45^\circ$,and $\alpha < \beta$. Then $\sum_{n=0}^\infty (\alpha^n + \frac{(-1)^n}{\beta^n})$ is equal to

Let $\alpha$ and $\beta$ be the roots of the quadratic equation $a x^2+b x+c=0$. Observe the lists given below:
List-$I$List-$II$
$(i)$ $\alpha = \beta$$(A)$ $(ac^2)^{1/3} + (a^2c)^{1/3} + b = 0$
$(ii)$ $\alpha = 2\beta$$(B)$ $2b^2 = 9ac$
$(iii)$ $\alpha = 3\beta$$(C)$ $b^2 = 6ac$
$(iv)$ $\alpha = \beta^2$$(D)$ $3b^2 = 16ac$
$(E)$ $b^2 = 4ac$
$(F)$ $(ac^2)^{1/3} + (a^2c)^{1/3} = b$

The correct match of List-$I$ from List-$II$ is:

The quadratic equation whose roots are $\sin ^2 18^{\circ}$ and $\cos ^2 36^{\circ}$ is

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo