Question

SOLVE THIS QUESTION FAST !!​

Answer 1

Answer:

$\huge \fbox \red {ANSWER࿐}$

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

Then,

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

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

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

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

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

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

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

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Answer 2

Answer:

(α+1)(β+1)=1−c

My answer is correct