Question

Plz help me ...
No. (vi) ...​

Answer 1

$\sf In \: quadratic \: equation, \: f(x) =ax^2+bx+c \\\\\sf Sum \: of \: roots \: (\alpha+\beta) = \frac{-b} {a} \\\\\sf Product \: of \: roots \:( \alpha\beta) = \frac{c} {a} \\\\\sf Now,\: In \: the \: question. \\\\\sf \frac{1}{a\alpha+b} + \frac{1} {a\beta+b} \\\\\sf For \: ease, \\\\\sf \: Let \: (a\alpha+b) \: be \: m \: and \: (a\beta+b) \: be \: n, \: then \\\\\sf Taking, \: LCM \: of \: denominators.. \\\\\sf We \: get, \\\\\sf \frac{1}{m}+ \frac{1}{n}=\frac{m+n} {mn} \\\\\sf Replace \: the \: values \: of \: m \: and \: n \\\\\sf \frac{(a\beta+b) +(a\alpha+b)} {(a\alpha+b) (a\beta+b)} \\\\\sf \frac{\orange{a}\beta+ \orange{a}\alpha+b+b} {a^2\alpha\beta + \green{ab}\alpha+\green{ab} \beta+b^2} \\\\\sf \frac{\orange{a}(\beta+\alpha) +2b} {a^2\alpha\beta + \green{ab}(\alpha+\beta) +b^2} \\\\\\\\\sf Putting \: the \: values \: of \: \alpha+\beta \: and \: \alpha\beta,\: we \: get \\\\\sf \frac{a \times \: (\frac{-b} {a} )+2b} {a^2 \times \: \frac{c} {a} + ab \times \: (\frac{-b} {a} )+b^2} \\\\\sf\frac{\cancel{a}\times \: (\frac{-b} {\cancel{a}}) +2b} {a \times \: \frac{c} {\cancel{a}} + \cancel{a}b\: x \:(\frac{-b} {\cancel{a}} )+b^2} \\\\\sf =\frac{b} {ac-b^2+b^2}=\frac{b} {ac- \cancel{b^2} +\cancel{b^2}}=\frac{b} {ac}$

Answer 2

$\begin{gathered} \sf In \: quadratic \: equation, \: f(x) =ax^2+bx+c \\\\\sf Sum \: of \: roots \: (\alpha+\beta) = \frac{-b} {a} \\\\\sf Product \: of \: roots \:( \alpha\beta) = \frac{c} {a} \\\\\sf Now,\: In \: the \: question. \\\\\sf \frac{1}{a\alpha+b} + \frac{1} {a\beta+b} \\\\\sf For \: ease, \\\\\sf \: Let \: (a\alpha+b) \: be \: m \: and \: (a\beta+b) \: be \: n, \: then \\\\\sf Taking, \: LCM \: of \: denominators.. \\\\\sf We \: get, \\\\\sf \frac{1}{m}+ \frac{1}{n}=\frac{m+n} {mn} \\\\\sf Replace \: the \: values \: of \: m \: and \: n \\\\\sf \frac{(a\beta+b) +(a\alpha+b)} {(a\alpha+b) (a\beta+b)} \\\\\sf \frac{\orange{a}\beta+ \orange{a}\alpha+b+b} {a^2\alpha\beta + \green{ab}\alpha+\green{ab} \beta+b^2} \\\\\sf \frac{\orange{a}(\beta+\alpha) +2b} {a^2\alpha\beta + \green{ab}(\alpha+\beta) +b^2} \\\\\\\\\sf Putting \: the \: values \: of \: \alpha+\beta \: and \: \alpha\beta,\: we \: get \\\\\sf \frac{a \times \: (\frac{-b} {a} )+2b} {a^2 \times \: \frac{c} {a} + ab \times \: (\frac{-b} {a} )+b^2} \\\\\sf\frac{\cancel{a}\times \: (\frac{-b} {\cancel{a}}) +2b} {a \times \: \frac{c} {\cancel{a}} + \cancel{a}b\: x \:(\frac{-b} {\cancel{a}} )+b^2} \\\\\sf =\frac{b} {ac-b^2+b^2}=\frac{b} {ac- \cancel{b^2} +\cancel{b^2}}=\frac{b} {ac}\end{gathered}</p><p>$