Question

If $\alpha \beta$ are the zeros of the polynomial $f(x)=x^{2}-p(x+1)-c,$ then $(\alpha +1)(\beta +1)$=
(a) c-1
(b) 1-c
(c) c
(d) 1+c

Answer 1

SOLUTION :

The correct option is (b) : 1 - c .

Given : α  and β are the zeroes of the  polynomial f(x) = x² -  p(x + 1) - c

f(x) = x² -  px - p - c

On comparing with ax² + bx + c,

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

Sum of the zeroes = −coefficient of x / coefficient of x²

α + β  = -b/a = - (- p)/1  

α + β  =  p ………………….(1)

Product of the zeroes = constant term/ Coefficient of x²

αβ = c/a = - p - c/1  

αβ = - p - c  ……………………(2)

Given : (α + 1)  (β + 1)

= αβ + β + α + 1

= αβ + (α + β )+ 1

= - p - c + p + 1

[From eq 1 & 2 ]

= - p + p - c + 1

= - c + 1

= 1 - c  

Hence, the value of (α + 1)(β + 1) is 1 - c .

HOPE THIS ANSWER WILL HELP YOU..

Answer 2

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

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

= x^2 + 1 + 2x - px -2p -c

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

As now alpha +1 and beta +1 are roots of f(x+1)

So product of roots = ( alpha +1)( beta +1)

= constant term/coefficient of x^2

= -2p - c +1