The $\alpha,\beta$ are the roots of the equation $x^{2} + ax + b = 0$, then $\frac{1}{\alpha^{2}} + \frac{1}{\beta^{2}}$ is equal to

Question

Dr. Abubacker · Accepted Answer

(a)
Here, $\alpha + \beta = - a,\ \ \ \ \alpha\beta = b$
$	herefore\ \ \frac{1}{\alpha^{2}} + \frac{1}{\beta^{2}} = \frac{\alpha^{2} + \beta^{2}}{(\alpha\beta)^{2}}$
$= \frac{(\alpha + \beta)^{2} - 2\alpha\beta}{(\alpha\beta)^{2}} = \frac{a^{2} - 2b}{b^{2}}$

Q. The $\alpha,\beta$ are the roots of the equation $x^{2} + ax + b = 0$, then $\frac{1}{\alpha^{2}} + \frac{1}{\beta^{2}}$ is equal to

Explanation

Reviewed by Dr. Abubacker

Want more practice?