Question

if a and b are zeroes of a quadratic polynomial x^2-p(x+1)-c show that (a+1) (b+1) = 1-c​

Answer 1

Step-by-step explanation:

use formula method

in x2-p(x+1)-c u get urs answer perfectly

Answer 2

Answer:

Hey friends!!

Here is your answer↓

it is given that

\alpha \: and \: \beta \: is \: the \: zeroes \: of \: polynomial \: f(x) = {x}^{2} - p(x + 1) - cαandβisthezeroesofpolynomialf(x)=x2−p(x+1)−c

alpha=œ

Beta=ß

f (x)=x²-px-p-c

=x²-px-(p+c)

By comparing quadratic equation ax²+bx+c

a=1; b=-p; c=-(p+c)

we know that

œ+ß=-b/a

=-(-p)/1 =-p

ϧ=c/a

=-(p+c)/1 =-(p+c)

A/Q,

L.H.S:-

=(œ+1)(ß+1)

=(œß)+(œ+ß)+1

Put the value of œ+ß and œß.

= -(p+c)+(-p)+1

= -p-c-p+1

= 1-c.

R.H.S :- 1-c.

Hence, it is proved L.H.S=R.H.S.

Hope it is helpful for you ✌✌