Question

if
$\alpha \: and \: \beta \: are \: the \: zeros \: of \: the \: quadratic \: polynomial \: p(y) = 5y{2}-7y -1 \: \: \: find \: the \: value \: of \: \binom{1}{ \alpha } + \binom{1}{ \beta }$
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Answer 1

$\star \bigstar\: {\huge \bold {\red{\mathfrak{ANSWER}}}} \: \: \bigstar \star$

$\star \: \: \underline \blue{given} \blue{ : }$

$5 {y}^{2} - 7y - 1 = 0 \\ = > {y}^{2} - \frac{7}{5} y - \frac{1}{5} = 0$

$\bold { \: \alpha \: and \: \beta \: are \: the \: zeros}$

$\star \: \: \underline \blue{find} \blue{ : }$

$\frac{1}{ \alpha } + \frac{1}{ \beta }$

$\star \: \: \underline \blue{solution} \blue{ : }$

We Know,

${\bold {\boxed {\mathfrak{ {x}^{2} - (sum \: of \: roots)x + product \: of \: roots = 0}}}}$

Hance,

$\alpha + \beta = \frac{7}{5} \\ \alpha \beta = - \frac{1}{5}$

$\frac{1}{ \alpha } + \frac{1}{ \beta } \\ \longrightarrow \: \frac{ \alpha + \beta }{ \alpha \beta } \\ \\ \bold{putting \: values \: of \: (\alpha + \beta ) \: and \: \alpha \beta }$

$\frac{ \frac{7}{5} }{ - \frac{1}{5} } \\ = > \boxed {\frac{1}{ \alpha } + \frac{1}{ \beta } = {- \frac{7}{2} }}$