Question

) If a and ß are zeroes of f(x)=x²- Px + q then find
1. a^2 + b^2
2. (1÷a) + (1÷b) ​

Answer 1

Answer:

$If \: \: \: \alpha \: \: \: and \: \: \: \beta \: \: \: \: are \: zeroes \: of \: \\ f(x) = {x}^{2} - p x + q \: \: \: \: \: then \: \\ \: \\ \alpha + \beta = - \frac{ - p}{1} = p \\ \alpha \beta = \frac{q}{1} = q \\ \\ (1) \\ \\ { \alpha }^{2} + { \beta }^{2} = {( \alpha + \beta )}^{2} - 2 \alpha \beta \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = {p}^{2} - 2q \\ \\ (2) \\ \\ \frac{1}{ \alpha } + \frac{1}{ \beta } \\ = \frac{ \alpha + \beta }{ \alpha \beta } \\ = \frac{p}{q}$

Answer 2

Answer:

1. p² - 2q

2. p/q

Step-by-step explanation:

refer to the attachment

Hope it helps !