Question

if a, bita are the roots of the equation x² - p(x+1) - C =0, then(a+1) (bita +1) =
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Answer 1

Answer:

Step-by-step explanation:

If α,β are the roots of x  

2

−p(x+1)+C=0⇒x  

2

−p x +C−p=0 then  

α+β=p and

α.β=C−p

Hence

(α+1)(β+1)  

=αβ+(α+β)+1

=C−p+(p)+1

=C+1.

Answer 2

Hii, let's Start.

$\rm\dag\large{\underline{\underline{Question:}}}$

if a, bita are the roots of the equation x² - p(x+1) - C =0, then(a+1) (bita +1) = ??

$\rm\dag\large{\underline{\underline{Given:}}}$

$\tt\large{ {x}^{2} \: - \: p(x + 1) \: - \: c \: } \: = \: 0$

$\rm\dag\large{\underline{\underline{Soluation:}}}$

$\tt\large{ { \: x}^{2} \: - \: p(x + 1) \: - \: c \: } = 0$

$\tt \longmapsto\large{ \: {x}^{2} \: - \: px \: - \: p \: - c \: = \: 0}$

$\tt\longmapsto\large{{x}^{2} \: - px \: - \: ( p + c) = \: 0 }$

$\rm \bf \bigstar\large \green{ \: \alpha + \beta \: = \: - \: p ( - \frac{ p}{1}) }$

$\rm \bf\rightarrow\large \green{ \alpha \: + \beta \: = p} \: \: \: \: \: \longrightarrow \: (i)$

$\rm\bf\bigstar\large \pink{ \: \alpha \beta \: = \: - ( { - \frac{p + c}{1}}) \: }$

$\rm\bf \rightarrow\large \pink{ \: \alpha \beta \: = - (p + c)} \: \: \: \longrightarrow(ii)$

$\rm\large{Now,we \: find \: ( \alpha + 1)( \beta + 1)}\: ,$

$\rm\large{ (\alpha + 1)( \beta + 1)}$

$\tt\large \longrightarrow{1 \: + \: \beta \: + \alpha \: + \alpha \beta}$

$\tt\large \longrightarrow{1 \: + \: p \: - (p + c)}$

$\tt\large \longrightarrow{1 \: + \: p \: - \: p \: - c}$

$\tt\large \longrightarrow \red{1 \: - \:C}$

$\rm{Hence, \:value \: of \:( \alpha + 1 ) \: ( \beta + 1) \: is \: 1 - C} \: .$

$\rule{300px}{ \: 4ex}$