Question

if alpha and beta are the zeroes of the polynomial x^2-p(x+1)-q. Find the value of alpha plus one into beta plus one.​

Answer 1

$\large\underline{\sf{Solution-}}$

Given that,

$\rm \: \alpha , \beta \: are \: zeroes \:of \: {x}^{2} - p(x + 1) - q$

can be rewritten as

$\rm \: \alpha , \beta \: are \: zeroes \:of \: {x}^{2} - px - p - q$

We know,

$\boxed{\red{\sf Sum\ of\ the\ zeroes=\frac{-coefficient\ of\ x}{coefficient\ of\ x^{2}}}}$

$\rm\implies \: \alpha + \beta = - \dfrac{( - p)}{1} = p$

Also,

$\boxed{\red{\sf Product\ of\ the\ zeroes=\frac{Constant}{coefficient\ of\ x^{2}}}}$

$\rm\implies \: \alpha \beta = \dfrac{ - p - q}{1} = - (p + q)$

Now, Consider

$\rm \: ( \alpha + 1)( \beta + 1)$

$\rm \: = \: \alpha ( \beta + 1) + 1( \beta + 1)$

$\rm \: = \: \alpha \beta + \alpha + \beta + 1$

So, on substituting the values, we get

$\rm \: = \: - (p + q) + p + 1$

$\rm \: = \: - p - q + p + 1$

$\rm \: = \: 1 - q$

Hence,

$\rm\implies \: \boxed{\tt{ \:\rm \: ( \alpha + 1)( \beta + 1) = 1 - q \: }}$

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

ADDITIONAL INFORMATION

$\red{\rm :\longmapsto\: \alpha , \beta , \gamma \: are \: zeroes \: of \: a {x}^{3} + b {x}^{2} + cx + d, \: then}$

$\boxed{ \bf{ \: \alpha + \beta + \gamma = - \dfrac{b}{a}}}$

$\boxed{ \bf{ \: \alpha \beta + \beta \gamma + \gamma \alpha = \dfrac{c}{a}}}$

$\boxed{ \bf{ \: \alpha \beta \gamma = - \dfrac{d}{a}}}$

Answer 2

Given that , α and β are zeroes of polynomial x²- p( x+ 1 ) -q .

Exigency To Find , The value of ( α + 1 ) ( β + 1 ) .

⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━⠀

• Polynomial :

$\longrightarrow \:\sf \:x^2\:- p( x+ 1 ) -q \:$

$\longrightarrow \:\sf \:x^2\:- px - p -q \:$

$\longrightarrow \:\sf \:x^2\:- px - ( p + q ) \:$

$\longrightarrow\purple{ \pmb{\:\sf \:x^2\:- px - ( p + q ) }}\:$

As We know that ,

$\dashrightarrow \sf \pink{\pmb{ \: \alpha \:+ \: \beta \:=\:-\dfrac{ Cofficient \:of\:x\:}{Cofficient \:of\:x^2\:}\:}}$

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

$\dashrightarrow \sf \: \alpha \:+ \: \beta \:=\:-\dfrac{( - p)\:}{1\:}\:$

$\dashrightarrow \purple{ \pmb{\:\sf \: \: \alpha \:+ \: \beta \:=\:p }}\:$

⠀⠀⠀⠀⠀⠀and,

$\dashrightarrow \sf \pink{\pmb{ \: \alpha \:\: \beta \:=\:\dfrac{ Constant \:term\:}{Cofficient \:of\:x^2\:}\:}}$

$\dashrightarrow \sf \: \alpha \: \: \beta \:=\:\dfrac{ Constant \:term\:}{Cofficient \:of\:x^2\:}\:$

$\dashrightarrow \sf \: \alpha \: \: \beta \:=\:\dfrac{-(p + q)\:}{1\:}\:$

$\dashrightarrow \purple{ \pmb{\:\sf \: \: \alpha \: \: \beta \:=\:-(p+ q ) }}\:$

⠀⠀⠀⠀⠀⠀⌬⠀Finding value of ( α + 1 ) ( β + 1 ) :

$\\\longrightarrow \sf \:(\:\alpha + 1 \:)\:\:(\:\beta + 1 \:)\:$

$\longrightarrow \sf \:\beta\:(\:\alpha + 1 \:)\:+ 1 \:(\:\alpha + 1 \:)\:$

$\longrightarrow \sf \:\alpha\beta + \beta \:+ \:\:\alpha + 1 \:\:$

$\longrightarrow \sf\:\alpha\beta + \alpha\:+ \:\:\beta + 1 \:$

$\longrightarrow \sf \:- p - q + p + 1 \:$

$\longrightarrow \sf\:- q + 1 \:$

$\longrightarrow \sf \: 1 - q \:$

$\longrightarrow\purple{ \pmb{\:\sf \: \:(\:\alpha + 1 \:)\:\:(\:\beta + 1 \:)\: \:=\: 1 - q }}\:$

$\:\:\:\:\:\therefore \:\sf \underline { \:Hence \: ,\:The\:Value \:of\: \:(\:\alpha + 1 \:)\:\:(\:\beta + 1 \:)\: is\:\pmb{ \:1 - q \:}\:.}$