Question

if a and b are the zeroes of the polynomial p (x)=x²-p(x+1)-c such that (a+1) (b+1)=O, the c =______​

Answer 1

Given -

• P(x) = x² - p(x + 1) - c

• (α + 1) (β + 1) = 0

To Find -

• Value of c

Formulae used -

• Sum of zeros

• Product of zeros

Solution -

In the question, we are provided with a polynomial an equation, that alpha + 1 and beta + 1 is equal to zero, and we need to find the value of c, for that, we will use 2 formulae, sum and product of zeros (alpha & beta) then we will put the values and then we will solve the further question.

General form of Quadratic equation -

→ ax² + bx + c = 0

So -

p(x) = x² - p(x + 1) - c

p(x) = x² - px - p - c

P(x) = x² - px - (p + c)

Where -

a = 1

b = -p

c = -(p + c)

Now -

→ Sum of zeros, α + β = -b/a

→ α + β = -(-p)/1

→ α + β = p

Similarly -

→ Product of zeros, αβ = c/a

→ αβ = -p/-c

→ αβ = -(p + c)/1

→ αβ = -(p + c)

On substituting the values -

→ (α + 1)(β + 1) = 0

→ αβ + α + β + 1 = 0

→ -p - c + p + 1 = 0

[ -p & +p will get cancelled]

→ -c + 1 = 0

→ - c = 0 - 1

→ - c = -1

→ c = 1

Therefore, the value of c is 1

______________________________

Answer 2

Given : $\alpha \:\:\& \beta \:\:$ are the zeroes of the polynomial p(x) = x² - p(x+1) - c

⠀⠀⠀&

⠀⠀⠀ $(\alpha + 1 ) \:(\beta + 1 ) \: = 0 \:\:$

Need To Find : The Value of c .

⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀

⠀⠀⠀⠀⠀Polynomial = p(x)=x²-p(x+1)-c

⠀⠀⠀First Simplify the Polynomial :

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

$\qquad :\implies \sf p(x)=x^2- px - p -c \;$

$\qquad :\implies \sf p(x)=x^2- px - ( p + c ) \;$

⠀⠀⠀⠀Therefore,

⠀⠀⠀⠀⠀Polynomial = p(x) = x²- px - ( p + c )

• Cofficient of x² = 1
• Cofficient of x = -p
• Constant Term = - ( p + c )

$\dag\:\:\it{ As,\:We\:know\:that\::}$

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

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

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

$\qquad :\implies \sf \alpha + \beta \:= \: \dfrac{p}{1} \;$

$\qquad :\implies \bf \alpha + \beta \:= \: p \;$

⠀⠀⠀⠀⠀&

$\dag\:\:\it{ As,\:We\:know\:that\::}$

$\qquad \dag\:\:\bigg\lgroup \sf{\:\:\: \alpha \beta\:\: \: \:: \dfrac{\:Constant\:Term\:}{Cofficient \:of\:x^2 }}\bigg\rgroup$

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

$\qquad :\implies \sf \alpha \beta \:= \: \dfrac{-(p+c) }{1} \;$

$\qquad :\implies \bf \alpha \beta \:= \: -(p+c) \;$

⠀⠀⠀⠀⠀Now ,

Given That ,

• $(\alpha + 1 ) \:(\beta + 1 ) \: = 0 \:\:$

⠀⠀⠀⠀⠀Now by Simplifying it :

$\qquad :\implies \sf (\alpha + 1 ) \:(\beta + 1 ) \: = 0 \;$

$\qquad :\implies \sf \alpha \:(\beta + 1 )\:+\:1 (\beta + 1 ) \: = 0 \;$

$\qquad :\implies \sf \alpha \beta + \alpha + \beta + 1 \: \: = 0 \;$

Here ,

• $\qquad :\implies \sf \alpha \beta \:= \: -(p+c) \;$
• $\qquad :\implies \sf \alpha + \beta \:= \: p \;$

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

$\qquad :\implies \sf -(p+c) + p + 1 \: \: = 0 \;$

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

$\qquad :\implies \sf \cancel{-p} - c \cancel{+ p} + 1 \: \: = 0 \;$

$\qquad :\implies \sf - c + 1 \: \: = 0 \;$

$\qquad :\implies \sf - c + 1 \: \: = 0 \;$

$\qquad :\implies \sf - c \: \: = -1 \;$ [ "-" sign will be eliminated from both side ]

$\qquad \longmapsto \frak{\underline{\purple{\:c\: =\:\: 1 \:\:}} }\bigstar$

Therefore,

⠀⠀⠀⠀⠀$\therefore {\underline{ \mathrm {\:The \:Value \:of\:c \:is\:\bf{1}}}}$

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