Question

If α andβ are zeroes of the polynomial x^2-p(x+1)+c such that (α+1)(β+1)=0, then find the value of c.

Answer 1

Answer:

c = -1

Step-by-step explanation:

x^2-p(x+1)+c

x^2 - px -p + c = 0

and α and β are the zeroes of the given polynomial

α+β = p      ------ (1)

αβ = c-p       ------(2)

therefore ,

(α+1)(β+1)=0

αβ + α + β + 1 = 0

put the value of α+β  and αβ

c - p + p + 1 = 0

c = -1

Answer 2

$\huge\mathfrak\blue{Answer:}$

Given:

• We have been given a Quadratic Polynomial x² - p ( x + 1 ) + c
• α and β are the zeros of given polynomial such that
• ( α + 1 )( β + 1 ) = 0

To Find:

• We have to find the value of c in the given Quadratic Polynomial

Solution:

We have been given a Polynomial

$\boxed{\text{f ( x ) = x² - p ( x + 1 ) + c}}$

$\text{f ( x ) = x² - px - p + c}$

$\text{f ( x ) = x² - px + ( c - p )}$

On Comparing Equation with Standard Form

$\boxed{\text{A = 1 and B = ( - p ) and C = c - p}}$

________________________________

$\odot \: \: \large{\underline{\text{Sum of zeros of Quadratic}}}$

$\mapsto$α + β $\sf{ = - \dfrac{B}{A} }$

$\mapsto$ α + β $\sf{ = - \left ( \dfrac{-p}{1} \right ) }$

$\mapsto$ $\boxed{ \text{α + β} \sf{ = p } }$

$\sf{ }$

$\odot \: \: \large{\underline{\text{Product of Zeros of Quadratic}}}$

$\mapsto$αβ $\sf{ = \dfrac{C}{A}}$

$\mapsto$ $\boxed{\text{αβ} \sf{ = c - p }}$

________________________________

$\underline{\large\mathfrak\red{According \: to \: the \: Question:}}$

$\implies$ ( α + 1 )( β + 1 ) = 0

$\implies$ αβ + ( α + β ) + 1 = 0

Substituting values in above Equation

$\implies$ c - p + p + 1 = 0

$\implies$ c + 1 = 0

$\implies$ c = -1

_________________________________

$\huge\underline{\sf{\red{A}\orange{n}\green{s}\pink{w}\blue{e}\purple{r}}}$

$\large\boxed{\sf{ Value \: of \: c = - 1}}$

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$\large\purple{\underline{\underline{\sf{Extra \: Information:}}}}$

• Polynomial : A Mathematical expression containing variables and coefficients involves Arithmetic operations
• Zeros of a Polynomial : Value of x for which the value of polynomial becomes zero.