Question

If α and β are the zeroes of the quadratic polynomial f(x) = x2 – p(x + 1) – c, show that (α + 1)(β + 1) = 1 – c.​

Answer 1

Step-by-step explanation:

Since, α and β are the zeroes of the quadratic polynomial

f(x) = x2 – p(x + 1)– c

Now,

Sum of the zeroes = α + β = p

Product of the zeroes = α × β = (- p – c)

So,

(α + 1)(β + 1)

= αβ + α + β + 1

= αβ + (α + β) + 1

= (− p – c) + p + 1

= 1 – c = RHS

So, LHS = RHS

Hence, proved.

Answer 2

----: Solution :----

As Given in QUESTION that 'α' and 'β' are the zeroes of Quadratic Polynomial

f(X) =x²-p(X+1)-c

And We have to Prove that

$(\alpha + 1 )( \beta + 1) = 1 - c$

Quadratic Equation = ax²+bx-c = 0

So,

$sum \: of \: the \: zeroes \: = > \alpha + \beta = p$

Product of the zeroes =α×β =–p–c

So,

$= ( \alpha + 1)( \beta + 1)$

$= \alpha ( \beta ) + ( \alpha + \beta ) + 1$

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

$= 1 - c = r.h.s$

Hence, L.H.S = R.H.S

1-c = 1-c

✔ proved ✔