Question

If alpha and beta are the zeros of quadratic polynomial p (x)=x2-k (x+1)-a evaluate (alpha +1)(beta+1)

Answer 1

Hi Mate !!


Given equation :- p ( x ) = x² - k ( x + 1 ) - a


x² - kx - k - a ..... ( Equation )


• Sum of the Zeros

$= \frac{ - coeff \: of \: {x} }{coeff \: of \: {x}^{2} }$

$\alpha + \beta = \frac{k}{1}$


• Product of Zeros :-

$= \frac{constant \: term \: }{coefficient \: of \: {x}^{2} }$


$\alpha \beta = \frac{ - k - a}{1}$

♯ To evaluate :-

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

$= \alpha \beta + \alpha + \beta + 1$

$= { - k - a} + k + 1$

$= 1 - a$

Answer 2

answer:

p(x)= x²-kx-k-p=0

a=1, b=-k, c=-k-p

∵α,β are zeroes of p(x)

∴α+β=-b/a⇒α+β=k

and αβ=c/a⇒αβ=-k-p

also(α+1)(β+1)=0

⇒αβ+α+β+1=0

⇒(-k-p)+k+1=0

              -p+1=0⇒p=1