Question

if alpha beta are the roots of x square + bx + c is zero then the value of 1 by a into alpha + beta + 1 by a into beta + b is​

Answer 1

Step-by-step explanation:

GIVEN :-

$\sf \: \alpha \: and \: \beta \: are \: the \: roots \: of \: the \: eq.$

$\sf \: {ax}^{2} + bx + c = 0$

TO FIND :-

$\sf \: \frac{1}{a \alpha + b} + \frac{1}{a \beta + b} =$

SOLUTION :-

$\sf \: Given \: \: Eq. -$

$\sf \: {ax}^{2} + bx + c = 0$

$\sf \: We \: \: know \: \: that -$

$\sf \: Sum \: \: of \: \: roots = ( \alpha + \beta ) = \frac{ - b}{a}$

$\sf \: Product \: \: of \: \: roots = ( \alpha \beta ) = \frac{c}{a}$

$\sf \: Now \: \: solving \: \: the \: \: sum \: \: of \: \: roots -$

$\sf \: \alpha + \beta = \frac{ - b}{a}$

$\implies \sf - b = a( \alpha + \beta )$

$\implies \sf b = - a( \alpha + \beta ) \: \: \: \: \: \: \: \: \: \: \: ......(i)$

$\sf \: According \: \: to \: \: Question :$

$\sf \: \frac{1}{a \alpha + b} + \frac{1}{a \beta + b}$

$= \sf \: \frac{1}{a \alpha - a( \alpha + \beta )} + \frac{1}{a \beta - a( \alpha + \beta )}$

$= \sf \: \frac{1}{a \alpha - a\alpha - a\beta } + \frac{1}{a \beta - a\alpha - a\beta }$

$= \sf \: \frac{1}{ \cancel{a \alpha} - \cancel{a\alpha} - a\beta } + \frac{1}{ \cancel{a \beta} - a\alpha - \cancel{ a\beta } }$

$= \sf \: \frac{1}{ - a \beta } + \frac{1}{ - a \alpha }$

$= \sf \: - \frac{1}{a} \bigg( \frac{1}{ \beta } + \frac{1}{ \alpha } \bigg)$

$\sf = - \frac{1}{a} \bigg( \frac{ \alpha + \beta }{ \alpha \beta } \bigg )$

$\sf \: Now \: \: putting \: \: the \: \: values -$

$\large\ \sf = - \frac{1}{a} \bigg( \frac{ \frac{ - b}{a} }{ \frac{c}{a} } \bigg )$

$= \sf \: - \frac{1}{a} \times \frac{ - b}{a} \times \frac{a}{c}$

$\sf \: = \frac{b}{ac}$

$\boxed{ \sf \: \frac{1}{a \alpha + b} + \frac{1}{a \beta + b} = \: \frac{b}{ac} }$