Question

if α and β are zeros of the polynomial x²-p(x+1)+c such that (α+1) (β+1) =0, then find the value of c.​

Answer 1

Given that , The α and β are zeros of the polynomial x²-p(x+1)+c such that (α+1) (β+1) =0 .

Exigency To Find : The value of c ?

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⠀⠀⠀⠀⠀⠀⠀G I V E N ⠀P O L Y N O M I A L : x² - p( x + 1 ) + c

$\qquad \dashrightarrow \sf Polynomial \:=\: x^2 - p ( x + 1 ) + c \:\:$

$\qquad \dashrightarrow \sf Polynomial \:=\: x^2 - p ( x + 1 ) + c \:\:$

$\qquad \dashrightarrow \sf Polynomial \:=\: x^2 - p x - p + c \:\:$

$\qquad \dashrightarrow \sf Polynomial \:=\: x^2 - p x + ( c - p ) \:\:$

$\qquad \dashrightarrow \sf Polynomial \:=\: x^2 - px + ( c - p ) \:\qquad \:\bigg\lgroup \sf{ Standard \:Form}\bigg\rgroup\:$

⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀

$\qquad \qquad \pmb{\mathbb{ POLYNOMIAL \:\::\:\: } \sf x^2 - px + ( c - p ) \:}$

As, We know that ,

$\qquad\underline {\boxed {\pmb{ \:\maltese \;Sum \:of \:zeroes \:\:\red {\:( \:\alpha \: + \beta )}\::}}}$

$\qquad \dashrightarrow \sf \bigg( \alpha \:+ \beta \: \bigg) \:=\:\dfrac{\:-(Cofficient \:of \:x\:)\:\:}{Cofficient \:of \:x^2 \:}$

⠀⠀⠀⠀⠀⠀$\underline {\boldsymbol{\star\:Now \: By \: Substituting \: the \: known \: Values \::}}$

$\qquad \dashrightarrow \sf \bigg( \alpha +\:\beta \: \bigg) \:=\:\dfrac{-(\:Cofficient \:of \:x\:)}{Cofficient \:of \:x^2 \:}\\\\\qquad \dashrightarrow \sf \bigg( \alpha \:+ \beta \: \bigg) \:=\:\dfrac{-(\:-p\:)}{1 \:}\\\\\qquad \dashrightarrow \sf \bigg( \alpha \:+ \beta \: \bigg) \:=\:\dfrac{p\:}{1 \:}\\\\\qquad \dashrightarrow \sf \alpha \:+ \beta \: \:=\:p\\\\\dashrightarrow \underline{\boxed{\purple{\pmb{\frak{ \:\alpha \:+ \beta \: \:=\:p\: }}}}}\:\:\bigstar$

⠀⠀⠀⠀⠀AND ,

$\qquad\underline {\boxed {\pmb{ \:\maltese \;Product \:of \:zeroes \:\:\red {\:( \:\alpha \: \beta )}\::}}}$

$\qquad \dashrightarrow \sf \bigg( \alpha \:\beta \: \bigg) \:=\:\dfrac{\:Constant \:Term\:\:}{Cofficient \:of \:x^2} \:$

⠀⠀⠀⠀⠀⠀$\underline {\boldsymbol{\star\:Now \: By \: Substituting \: the \: known \: Values \::}}$

$\qquad \dashrightarrow \sf \bigg( \alpha \:\beta \: \bigg) \:=\:\dfrac{\:Constant\:Term\:}{Cofficient \:of \:x^2 \:}\\\\\qquad \dashrightarrow \sf \bigg( \alpha \: \beta \: \bigg) \:=\:\dfrac{\:( c - p )\:}{1 \:}\\\\\qquad \dashrightarrow \sf \bigg( \alpha \: \beta \: \bigg) \:=\:\cancel{\dfrac{\:( c - p )\:}{1 \:}}\\\\\qquad \dashrightarrow \sf \bigg( \alpha \: \beta \: \bigg) \:=\:( c - p ) \\\\ \dashrightarrow \underline{\boxed{\purple{\pmb{\frak{ \:\alpha \: \beta \: \:=\:( c - p )\: }}}}}\:\:\bigstar$

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Given that ,

• (α+1) (β+1) =0

$\qquad \dashrightarrow \sf \Big\{ \alpha + 1 \Big\} \: \Big\{ \beta + 1 \Big\} \: = \:0 \:$

$\qquad \dashrightarrow \sf \Big\{ \alpha + 1 \Big\} \: \Big\{ \beta + 1 \Big\} \: = \:0 \:$

$\qquad \dashrightarrow \sf \alpha \Big\{ \beta + 1 \Big\} \:+ 1 \Big\{ \beta + 1 \Big\} \: = \:0 \:$

$\qquad \dashrightarrow \sf \alpha\beta + \alpha \:+ \beta + 1 \: = \:0 \:$

$\qquad \dashrightarrow \sf \Big\{ \alpha\beta \Big\} + \Big\{ \alpha \:+ \beta \Big\} + 1 \: = \:0 \:$

⠀⠀⠀⠀⠀⠀$\underline {\boldsymbol{\star\:Now \: By \: Substituting \: the \: known \: Values \::}}$

$\qquad \dashrightarrow \sf \Big\{ \alpha\beta \Big\} + \Big\{ \alpha \:+ \beta \Big\} + 1 \: = \:0 \:$

$\qquad \dashrightarrow \sf \Big\{ c - p \Big\} + \Big\{ p \Big\} + 1 \: = \:0 \:$

$\qquad \dashrightarrow \sf c - p + p + 1 \: = \:0 \:$

$\qquad \dashrightarrow \sf c + 1 \: = \:0 \:$

$\qquad \dashrightarrow \sf c \: = \:-1 \:$

$\dashrightarrow \underline{\boxed{\purple{\pmb{\frak{ \:c \: \:=\:-1\: }}}}}\:\:\bigstar$

$\qquad \therefore \:\underline {\sf Hence, \:The \:Value \:of \:c \: is \:\pmb{\bf - 1 \:}\:\:.}$