Question

Pls answer me fast not abbusive answer​

Answer 1

Step-by-step explanation:

$(i) \: p {x}^{2} - q \: \\ equating \: with \: a {x}^{2} + bx + c \: we \: \\ find \: \\ a = p \\ b = 0 \\ c = - q \\ let \: \alpha \: and \: \beta \: be \: the \: be \: the \: zeros \\ hence \: sum \: of \: zeros \\ \alpha + \beta \: = - \frac{b}{a} = \frac{0}{p} = 0 \\ product \: of \: zeros \: \\ \alpha \beta = \frac{c}{a} = \frac{ - q}{p} = - \frac{q}{p} \\ \\ (ii)\: q {x}^{2} + q \: \\ equating \: with \: a {x}^{2} + bx + c \: we \: \\ find \: \\ a = q \\ b = 0 \\ c = q \\ let \: \alpha \: and \: \beta \: be \: the \: be \: the \: zeros \\ hence \: sum \: of \: zeros \\ \alpha + \beta \: = - \frac{b}{a} = \frac{0}{q} = 0 \\ product \: of \: zeros \: \\ \alpha \beta = \frac{c}{a} = \frac{ q}{q} = 1$