Question

If α and β are the zeroes of the polynomial x2-p(x+1)-c, then (α+1)(β+1) equals to what

Answer 1

Given

α and β are the zeroes of the polynomial x2-p(x+1)-c, then (α+1)(β+1) ?

Solution :-

$\sf\ x^2-px-p-c=0 \\ \\ \\ \implies\sf\ x^2-px-(p+c) =0$

here -

$\implies\sf\ \alpha+\beta= \dfrac{-b}{a}\\ \\ \implies\sf\ \alpha\beta=\dfrac{c}{a}\\ \\ \sf\ we \ have \ a= 1\ \; \ b= -p\ and \ c= -(p+c)\\ \\ \longrightarrow \sf\ \alpha+\beta= \dfrac{-(-p)}{1}\\ \\ \longrightarrow\sf\ \alpha+\beta= p\\ \\ \\ \longrightarrow\sf\ \alpha\beta= \dfrac{-(p+c)}{1}\\ \\ \\ \longrightarrow\sf\ \alpha\beta= -(p+c)$

Now ,

$\sf\ (\alpha+1)(\beta+1)\\ \\ \implies\sf\ \alpha\beta+\alpha+\beta+1 \\ \\ \implies\sf\ \alpha\beta+(\alpha+\beta)+1\\ \\ \\ \sf\ by\ putting \ the \ values \\ \\ \\ \implies\sf\ -(p+c)+p+1\\ \\ \\ \implies\sf\ \cancel{-p}-c+\cancel{p}+1\\ \\ \\ \implies\sf\ (\alpha+1)(\beta+1)=1-c$

Answer 2

$\huge{\boxed{\bold{Question}}}$

If α and β are the zeroes of the polynomial x2-p(x+1)-c, then (α+1)(β+1) equals to what

Answer

c= -1

step by step explanation

→ x²-p(x-1)+c

→ x²-px-p+c

→ x²-px+(c-p)

Comparing with ax²+bx+c we get

a=1

b=-p

c=c-p

Given

(α+1)(β+1)=0

αβ+α+β+1=0

Note that sum of roots=-b/a

α+β=-b/a

But b=-p

a=1

so, α+β=-(-p)/1=p

product of roots = αβ=c/a

=αβ=(c-p)

Hence write this as

αβ +α+β+1=0

c-p+p+1=0

c+1=0

c=-1