If $\alpha, \beta$ are the roots of $x^2 - px + q = 0$ and $\alpha', \beta'$ are the roots of $x^2 - p'x + q' = 0$,then the value of $(\alpha - \alpha')^2 + (\beta - \alpha')^2 + (\alpha - \beta')^2 + (\beta - \beta')^2$ is

  • A
    $2\{p^2 - 2q + p'^2 - 2q' - pp'\}$
  • B
    $2\{p^2 - 2q + p'^2 - 2q' - qq'\}$
  • C
    $2\{p^2 - 2q - p'^2 - 2q' - pp'\}$
  • D
    $2\{p^2 - 2q - p'^2 - 2q' - qq'\}$

Explore More

Similar Questions

If $\alpha$ and $\beta$ are roots of the equation $x^2 + px + \frac{3p}{4} = 0,$ such that $|\alpha - \beta| = \sqrt{10},$ then $p$ belongs to the set

Difficult
View Solution

If $A.M.$ of the roots of a quadratic equation is $8/5$ and $A.M.$ of their reciprocals is $8/7$,then the equation is

If $a, b, c$ are distinct real numbers and $a^3 + b^3 + c^3 = 3abc$,then the equation $ax^2 + bx + c = 0$ has two roots,out of which one root is

Difficult
View Solution

If exactly one root of the equation $x^2 + (a - 1)x + 2a = 0$ lies in the interval $(0, 3)$,then the set of values of $a$ is given by:

Difficult
View Solution

If $x = \sqrt{5} + 2$,then the value of $\frac{2x^2 - 3x - 2}{3x^2 - 4x - 3}$ is equal to

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