Question

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Answer 1

Answer:

$\bold {Since \: \alpha \: and \: \ \beta \: are \: the \: zeros \: of \: quadratic \: polynomial \: f (x) = {x}^{2} - p \: (x + 1) - c}$

Then,

$= \bf {x}^{2} - p(x + 1) - c \\ = \bf {x}^{2} - px - p - c \\ = \bf \alpha + \beta = \frac{ - coefficient \: of \: x}{coefficient \: of \: {x}^{2} } \\ = \bf - \ \frac{ - ( - p)}{1} = \\ = \bf \red p$

$\bf\alpha \beta = \frac{constant \: term}{coefficient \: of \: {x}^{2} } \\ \bf = - \frac{ - p - c}{1} \\ \bf \purple {= - p - c}$

$\bold {we \: have \: to \: prove \: that \: ( \alpha + 1)( \beta + 1) = 1 - c}$

=(α+1)(β+1)=1−c

=(α+1)β+(α+1)(1)=1−c

=αβ=β+α+1=1−c

=αβ+(α+β)+1=1−c

$\begin{gathered} \bold {Substituting \: \alpha + \beta = p \: and \: \alpha \beta = - p - c} \\ \bold {We \: get} : \end{gathered}$

We get:

$\begin{gathered} \bf = - p - c + p + 1 = 1 - c \\ \bf 1 - c = 1 - c\end{gathered}$

$\bold {Hence \: its \: shown \: that \: (\alpha + 1)( \beta + 1) = 1 - c}$

Answer 2

Answer:

\bold {Since \: \alpha \: and \: \ \beta \: are \: the \: zeros \: of \: quadratic \: polynomial \: f (x) = {x}^{2} - p \: (x + 1) - c}Sinceαand βarethezerosofquadraticpolynomialf(x)=x

2

−p(x+1)−c

Then,

\begin{gathered} = \bf {x}^{2} - p(x + 1) - c \\ = \bf {x}^{2} - px - p - c \\ = \bf \alpha + \beta = \frac{ - coefficient \: of \: x}{coefficient \: of \: {x}^{2} } \\ = \bf - \ \frac{ - ( - p)}{1} = \\ = \bf \red p\end{gathered}

=x

2

−p(x+1)−c

=x

2

−px−p−c

=α+β=

coefficientofx

2

−coefficientofx

=−

1

−(−p)

=

=p

\begin{gathered} \bf\alpha \beta = \frac{constant \: term}{coefficient \: of \: {x}^{2} } \\ \bf = - \frac{ - p - c}{1} \\ \bf \purple {= - p - c} \end{gathered}

αβ=

coefficientofx

2

constantterm

=−

1

−p−c

=−p−c

\bold {we \: have \: to \: prove \: that \: ( \alpha + 1)( \beta + 1) = 1 - c}wehavetoprovethat(α+1)(β+1)=1−c

=(α+1)(β+1)=1−c

=(α+1)β+(α+1)(1)=1−c

=αβ=β+α+1=1−c

=αβ+(α+β)+1=1−c

\begin{gathered} \begin{gathered} \bold {Substituting \: \alpha + \beta = p \: and \: \alpha \beta = - p - c} \\ \bold {We \: get} : \end{gathered} \end{gathered}

Substitutingα+β=pandαβ=−p−c

Weget:

We get:

\begin{gathered}\begin{gathered} \bf = - p - c + p + 1 = 1 - c \\ \bf 1 - c = 1 - c\end{gathered} \end{gathered}

=−p−c+p+1=1−c

1−c=1−c

\bold {Hence \: its \: shown \: that \: (\alpha + 1)( \beta + 1) = 1 - c} Henceitsshownthat(α+1)(β+1)=1−c