Question

If \(\alpha,\beta\) are the roots of \(ax^2+bx+c=0\), then \[ \left(\frac{\alpha}{\beta}-\frac{\beta}{\alpha}\right)^2 = \]

• A. \(\dfrac{b^2(b^2-4ac)}{a^2c^2}\)
• B. \(\dfrac{b^2(b^2-4ac)}{a^4}\)
• C. \(\dfrac{b^2(b^2-4ac)}{ca^3}\)
• D. \(\dfrac{b^2(b^2-4ac)}{c^4}\)

Hint:

Useful relations: \[ \boxed{ \alpha+\beta=-\frac{b}{a},\qquad \alpha\beta=\frac{c}{a} } \]

Answer 1

The Correct Option is A

Using the relations between roots, \[ \alpha+\beta=-\frac{b}{a},\qquad \alpha\beta=\frac{c}{a}. \] Now, \[ \left(\frac{\alpha}{\beta}-\frac{\beta}{\alpha}\right)^2 = \frac{(\alpha^2-\beta^2)^2}{(\alpha\beta)^2}. \] Since \[ \alpha^2-\beta^2=(\alpha+\beta)(\alpha-\beta), \] and \[ (\alpha-\beta)^2 =(\alpha+\beta)^2-4\alpha\beta = \frac{b^2-4ac}{a^2}, \] therefore, \[ (\alpha^2-\beta^2)^2 = \frac{b^2(b^2-4ac)}{a^4}. \] Also, \[ (\alpha\beta)^2=\frac{c^2}{a^2}. \] Hence, \[ \left(\frac{\alpha}{\beta}-\frac{\beta}{\alpha}\right)^2 = \frac{b^2(b^2-4ac)}{a^2c^2}. \] Therefore, \[ \boxed{(A)} \]