Question

If alpha and beta are zeroes of the polynomial x2-p(x+1)+c such that (alpha+1)(beta+1)=0 , then find the value of c.

Answer 1

x² - p(x+1)-c = x² - px-p -c compare it with ax²+bx+c =0 a= 1, b= -p , c= -p-c α and β are two zeroes

Answer 2

Answer:

HOPE IT HELP YOU

Step-by-step explanation:

x² - p ( x + 1 ) + c

⇒ x² - p x - p + c

⇒ x² - p x + ( 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

Hence, the value of c is - 1.