1.

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

Answer»

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


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