Question

if alpha and beta are the zeros of the quadratic polynomial f(x)=x^-p(x+1)-c, show that (alpha+1)(beta+1)=1-c

Answer 1

Hey!

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Given,

Alpha and beta are the zeroes of the quadratic polynomial x^2 - p (x + 1) - c

Alpha ( @ ) 
Beta ( ß )


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

Comparing with ax^2 + bx + c

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

We know,

Alpha + Beta = - b/a = - (-p) = p
Alpha × Beta = c/a = - ( p+c ) 

Thus,

( Alpha + 1 ) ( Beta + 1 ) 
= @ß + @ + ß + 1
= - (p+c) + p + 1
= - p - c + p + 1
= 1 - c

Thus, ( Alpha + 1 ) ( Beta + 1 ) = 1 - c


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Hope it helps...!!!

Answer 2

Hey friends!!

Here is your answer↓

it is given that
$\alpha \: and \: \beta \: is \: the \: zeroes \: of \: polynomial \: f(x) = {x}^{2} - p(x + 1) - c$
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 ✌✌✌.