Question

plz solve fast and give anwser

Answer 1

$\underline{\mathfrak{Solution : }}$

$\underline{\mathsf{Given, }}$

$\mathsf{ \implies p(x) \: = \: {x}^{2} \: - \: k(x \: + \: 1) \: - \: a} \\ \\ \mathsf{ \implies p(x) \: = \: {x}^{2} \: - \: kx \: - \: k \: - \: a} \\ \\ \mathsf{ \implies p(x) \: = \: {x}^{2} \: - \: kx \: - \: (k \: + \: a)}$

$\underline{\mathsf{Here, }}$

$\mathsf{\implies Coefficient \: of \: {x}^{2} \: = \: 1} \\ \\ \mathsf{\implies Coefficient \: of \: x \: = \: -k } \\ \\ \mathsf{\implies Constant \: term \: = \: -(k \: + \:a )}$

$\underline{\mathsf{And, }}$

$\mathsf{\alpha \: and \: \beta \: are \: zeroes \: of \: p(x).}$

$\mathsf{According \: to \: the \: question, } \\ \\ \mathsf{\implies (\alpha \: + \: 1)(\beta \: + \: 1 ) \: = \: 0} \\ \\ \mathsf{\implies \alpha \beta \: + \: \alpha \: + \: \beta \: + \: 1 \: = \: 0 } \\ \\ \mathsf{ \implies ( \alpha \: + \: \beta) \: + \: \alpha \beta \: + \: 1 \: = \: 0 \qquad...(1)}$

$\mathsf{We \: know \: the \: relationship \: between \: coeffi-} \\ \mathsf{cient \: of \: a \: quadratic \: polynomial \: and \: its} \\ \mathsf{ zeroes. }$

$\mathsf{\implies Sum \: of \: zeroes \: = \: -\dfrac{Coefficient \: of \: x}{Coefficient \: of \: {x}^{2}}}$

$\mathsf{\implies \alpha \: + \: \beta \: = \: -\dfrac{(-k)\:}{1} } \\ \\ \mathsf{\implies \alpha \: + \: \beta \: = \: k \qquad...(2) }$

$\underline{\mathsf{And,}} \\ \\ \mathsf{\implies Product \: of \: zeroes \: = \: \dfrac{Constant \: term}{Coefficient \: of \: {x}^{2}}} \\ \\ \mathsf{\implies \alpha \beta \: = \: \dfrac{-(k \: + \: a )\:}{1}} \\ \\ \mathsf{\implies \alpha \beta \: = \: -k \: - \:a \qquad...(3)}$

$\mathsf{Plug \: the \: value \: of \: (2) \: and \: (3) \: in \: (1),} \\ \\ \mathsf{\implies (\alpha \: + \: \beta) \: + \: \alpha \beta \: + \: 1 \: = \: 0} \\ \\ \mathsf{\implies \cancel{ k }\: - \: \cancel{k} \: - \: a \: + \: 1 \: = \: 0} \\ \\ \mathsf{ \implies \cancel{ - }a \: = \: \cancel{ - }1} \\ \\ \mathsf{ \therefore \quad \: a \: = \: 1}$

$\boxed{\underline{\mathsf{The \: value \: of \: a \: is \: 1.}}}$