Question

Hello, in the pic, can someone please solve the 6th subdivision of Q 20.
Thanks : )

Answer 1

Step-by-step explanation:

We have,

$\tt \: f(x) = a {x}^{2} + bx + c$

It's zeros are $\tt\blue{\alpha\:\:\&\:\:\beta}$

So,

$\tt \large{\purple{ • \: Sum \: \: of \: \: roots \: \: , (\alpha + \beta ) = - \frac{b}{a} }}$

$\tt \large{\purple{ • \: Product\: \: of \: \: roots \: \: , (\alpha \beta ) = \frac{c}{a} }}$

Now,

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

$\sf \color{violet} = \frac{a \beta + b + a \alpha + b}{(a \alpha + b)(a \beta + b)}$

$\sf \color{violet} = \frac{a( \alpha + \beta ) + 2b }{a ^{2} \alpha \beta + ab \beta + ab \alpha + b^{2} }$

$\sf \color{violet} = \frac{a( \alpha + \beta ) + 2b }{a ^{2} \alpha \beta + ab (\alpha + \beta ) + b^{2} }$

$\sf \color{violet} = \frac{a \bigg( - \dfrac{b}{a} \bigg ) + 2b }{a ^{2}\cdot \dfrac{c}{a} + ab \bigg( - \dfrac{b}{a} \bigg) + b^{2} }$

$\sf \color{violet} = \frac{ - b + 2b }{ac - b^{2} + b^{2} }$

$\sf \color{violet} = \frac{ b }{ac }$