Question

If α , β are the roots of the quadratic equation
x² - p(x + 1) - c = 0 then (α + 1)(β + 1) =

a) c - 1
b) 1 - c
c) c
d) 1 + c​

Answer 1

Question :

If α , β are the roots of the quadratic equation x² - p(x + 1) - c = 0 then (α + 1)(β + 1) ?

Solution :

We have ,

$\sf\:f(x)={x}^{2}-p(x+1)-c$

$\sf \implies \: f(x) = x{}^{2} - px - p- c$

$\sf \implies \: f(x) = {x}^{2} - px - (p + c)$

We know that ,

$\sf \alpha + \beta = \dfrac{ - cofficient \: of \: x}{cofficient \: of \: x {}^{2} }$

$\sf \implies \alpha + \beta = \dfrac{ - ( - p)}{ 1} = p...(1)$

$\sf \: and \: \alpha \beta = \dfrac{constant}{cofficient \: of \: x {}^{2} }$

$\sf \implies \alpha \beta = \dfrac{ - (p + c)}{1} = - (p + c)..(2)$

We have to find the value of

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

$\sf = \alpha \beta + \alpha + \beta + 1$

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

Now put the values of equation (1) and (2)

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

$\sf = - p - c + p + 1$

$\sf = 1 - c$

Correct option b)

Answer 2

Answer:

If α , β are the roots of the quadratic equation

x² - p(x + 1) - c = 0 then (α + 1)(β + 1) =

a) c - 1

b) 1 - c

c) c

d) 1 + c