Question

If alpha and Beta are zeroes of the polynomial x² -p (x +1) + c such that (alpha + 1)( beta+1)= 0, then find the value of c.​

Answer 1

Answer:

We have,

$\:$

• f(x) = x² - p(x + 1) - c = 0

$\:$

• f(x) = x² - px - (p + c) = 0

$\:$

Since,

$\:$

$\large \mathtt{ \alpha \: \: \beta \: \: are \: \: zeros \: \: of \: \: p(x)}$

$\:$

So,

$\:$

$\large \mathrm{ \alpha + \beta = p}$

$\:$

$\large : \implies \: \mathrm{ \alpha \beta = - (p + c)}$

$\:$

Since,

$\:$

$\large \mathtt{( \alpha + 1)( \beta + 1) = 0}$

$\:$

$\large \: \: \: \: \: : \implies \mathtt{ \alpha \beta + \alpha + \beta + 1= 0}$

$\:$

$\large \: \: \: \: \: : \implies \mathtt{ - p - c + p + 1= 0}$

$\:$

$\large \: \: \: \: \: : \implies \mathtt{ \cancel{ - p }- c + \cancel p + 1= 0}$

$\:$

$\large \: \: \: \: \: : \implies \mathtt{ - c + 1= 0}$

$\:$

$\large \boxed{ \rm \: \therefore \: c = 1}$

Answer 2

Answer:

We have,

$\:$

• f(x) = x² - p(x + 1) - c = 0

$\:$

• f(x) = x² - px - (p + c) = 0

$\:$

Since,

$\:$

$\large \mathtt{ \alpha \: \: \beta \: \: are \: \: zeros \: \: of \: \: p(x)}$

$\:$

So,

$\:$

$\large \mathrm{ \alpha + \beta = p}$

$\:$

$\large : \implies \: \mathrm{ \alpha \beta = - (p + c)}$

$\:$

Since,

$\:$

$\large \mathtt{( \alpha + 1)( \beta + 1) = 0}$

$\:$

$\large \: \: \: \: \: : \implies \mathtt{ \alpha \beta + \alpha + \beta + 1= 0}$

$\:$

$\large \: \: \: \: \: : \implies \mathtt{ - p - c + p + 1= 0}$

$\:$

$\large \: \: \: \: \: : \implies \mathtt{ \cancel{ - p }- c + \cancel p + 1= 0}$

$\:$

$\large \: \: \: \: \: : \implies \mathtt{ - c + 1= 0}$

$\:$

$\large \boxed{ \rm \: \therefore \: c = 1}$