Question

For p(x) = ax2+bx+c then find 1/alpha2+ 1/beta2, alpha , beta are the zeros of p(x)​

Answer 1

Answer:

Let alpha = A and beta = B

$a {x}^{2} + bx \: + c$

So,

A + B = -b/c

A × B = c/a

1/A + 1/B

= A + B / AB

= -b/c ÷ c/a

= - b/c × a/c

= -ab / c^2