Question

If alpha and beta are the zeroes of polynomial p(X)=xsquare-p(X+1)c such that (alpha +1) (beta+1)=0. Find c

Answer 1

Answer:

Step-by-step explanation:Given that alpha and beta are the roots of the quadratic equation f(x) = x^2-p(x+1)-c = x^2-px-p-c = x^2 -px-(p+c),

comparing with ax^2 + bx + c, we have, a =1 , b= -p & c= -(p+c)

alpha+beta = -b/a = -(-p)/1 = p

& alpha*beta = c/a = -(p+c)/1 = -(p+c)

Therefore, (Alpha + 1)*(beta+1)

= Alpha*beta + alpha + beta + 1

= -(p+c) + p + 1

= -p-c+p+1

= 1-c

or c=1

Answer 2

Correct question : If α and β are the zeros of the polynomial x² - p ( x + 1 ) + c such that, ( α + 1 ) ( β + 1 ) = 0 then find the value of c.

Answer:

c = - 1

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