Question

pls someone give the right answer​

Answer 1

Answer:

hope this will help you

b) is your answer....

Please mark me as brainliest ♥️♥️♥️

because this answer deserves it....

may god bless you.....

Answer 2

As $\alpha$ and $\beta$ are the roots of $\sf 3x^2 + 5x-7=0$, so,

$\sf \alpha + \beta = \frac{ - b}{a} = \frac{ - 5}{3} \quad\quad\dots(1) \\ \\ \longrightarrow \sf 3(\alpha + \beta ) = - 5 \\ \\ \longrightarrow \sf 3 \alpha + 3 \beta = - 5 \\ \\ \longrightarrow \sf 3 \alpha + 5 = - 3 \beta \quad\quad\dots(2) \\ \\ \longrightarrow \sf 3 \beta + 5 = - 3 \alpha\quad\quad\dots(3)$

And,

$\sf \alpha \beta = \dfrac{c}{a} = \dfrac{ - 7}{3}\quad\quad\dots(4)$

And,

$\sf {( \alpha + \beta )}^{2} = { \alpha }^{2} + { \beta }^{2} + (2 \alpha \beta)$

From (1) and (4),

$\\ \longrightarrow \sf { \alpha }^{2} + { \beta }^{2} = {\bigg( \dfrac{ - 5}{3}\bigg) }^{2} - \bigg(2 \times \dfrac{ - 7}{3} \bigg) \\ \\ \longrightarrow \sf { \alpha }^{2} + { \beta }^{2} = \dfrac{25}{9} + \dfrac{14}{3} \\ \\ \longrightarrow \sf { \alpha }^{2} + { \beta }^{2} = \dfrac{67}{9} \quad\quad\dots(5)$

To find:

$\sf {\bigg( \dfrac{1}{3 \alpha + 5}\bigg) }^{2} +{\bigg( \frac{1}{3 \beta + 5}\bigg) }^{2}$

From (2) and (3),

$\\ \\ \longrightarrow \sf {\bigg( \frac{1}{ - 3 \beta }\bigg)}^{2} + {\bigg( \frac{1}{ - 3 \alpha }\bigg) }^{2} \\ \\ \longrightarrow \sf \frac{1}{9 { \beta }^{2} } + \frac{1}{9 { \alpha }^{2} } \\ \\ \longrightarrow \sf \frac{1}{9} \bigg( \frac{1}{ { \beta }^{2} } + \frac{1}{ { \alpha }^{2} } \bigg) \\ \\ \longrightarrow \sf \frac{1}{9} \bigg( \frac{ { \alpha }^{2} + { \beta }^{2} }{ {( \alpha \beta )}^{2} } \bigg)$

From (5) and (4),

$\\ \\ \longrightarrow \sf \frac{1}{9} \Bigg[ \frac{ \big(\frac{67}{9}\big) }{ \big({ \frac{ - 7}{3} }^{2} \big)} \Bigg] \\ \\ \longrightarrow \sf \frac{1}{9} \Bigg[ \frac{\:\: \big(\frac{67}{9}\big)\:\:}{ \:\:\big(\frac{49}{9}\big) \:\:} \Bigg] \\ \\ \longrightarrow \sf \frac{1}{9} \bigg( \frac{67}{49} \bigg) \\ \\ \longrightarrow \sf \dfrac{67}{441 \: }$

Hence the answer is,

$\longrightarrow \underline{\underline{\sf{\green{ \:\:\dfrac{67}{441}\:\:}}}}$