Question

if p(x)= cx^2+dx-c. Find alpha square+beta square​

Answer 1

$\underline\mathfrak{Given:-}$

• $\: \: \: \: \: \: \: p \: {(x)} \: \: \leadsto \: \: {cx}^{2} \: + \: {dx} \: - \: {c}$

$\underline\mathfrak{To \: \: Find:-}$

• $\: \: \: \: \: Value \: \: of \: \: {\alpha}^{2} \: + \: {\beta}^{2}?$

$\underline\mathfrak{Solutions:-}$

• $\: \: \: \: \: \: \: p \: {(x)} \: \: \leadsto \: \: {cx}^{2} \: + \: {dx} \: - \: {c}$

$\: \: \: \: \: \leadsto {a} \: \: = \: \: {c}$

$\: \: \: \: \: \leadsto {b} \: \: = \: \: {d}$

$\: \: \: \: \: \leadsto {c} \: \: = \: \: {-c}$

• $\: \: \: \: \: {Let \: \: \alpha \: \: and \: \: \beta \: \: are \: \: the \: \: zeroes \: \: of \: \: the \: \: given \: \: polynomial}$

$\: \: \: \: \: \leadsto {\alpha} \: + \: {\beta} \: \: = \: \: \dfrac{-b}{a}$

$\: \: \: \: \: \leadsto \dfrac{-d}{c}$

$\: \: \: \: \: \leadsto {\alpha\beta} \: \: = \: \: \dfrac{c}{a}$

$\: \: \: \: \: \leadsto \dfrac{-c}{c}$

$\: \: \: \: \: \leadsto {-1}$

$\: \: \: \: \: {({\alpha} \: + \: {\beta})}^{2} \: \: \leadsto \: \: {\alpha}^{2} \: + \: {\beta}^{2} \: + \: {{2}\alpha\beta}$

$\: \: \: \: \: \leadsto {({\alpha} \: + \: {\beta})}^{2} \: + \: {{2}\alpha\beta}$

• $\: \: \: \: \: put \: \: the \: \: of \: \: {\alpha} \: + \: {\beta} \: \: and \: \: {\alpha\beta}$

$\: \: \: \: \: \leadsto {(\dfrac{-d}{c})}^{2} \: + \: {{2} \: \times \: {-1}}$

$\: \: \: \: \: \leadsto {\dfrac{d}{{c}^{2}}}^{2} \: + \: {{(-2)} \: \times \: {-1}}$

$\: \: \: \: \: \leadsto {\dfrac{{d}^{2}}{{c}^{2}}^{2} \: + \: {2}}$

$\: \: \: \: \: \leadsto \dfrac{{d}^{2} \: + \: {2c}^{2}}{{c}^{2}}$

• $\: \: \: \: \: So, \: \: Value \: \: of \: \: {\alpha}^{2} \: + \: {\beta}^{2} \: \: is \: \: \dfrac{{d}^{2} \: + \: {2c}^{2}}{{c}^{2}}$

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Answer 2

Answer:

d²+2c²/c² is your answer