Question

If a, b are roots of equation 7x^2 — 3x + 2 = 0 then find the value of

Answer 1

Step-by-step explanation:

is it correct? kindly reply for it.

Answer 2

ANSWER.

$\sf \to \: \dfrac{ \alpha }{1 - \alpha {}^{2} } + \dfrac{ \beta }{1 - { \beta }^{2} } = \dfrac{ - 21}{16}$

EXPLANATION.

$\sf \to \: \alpha \: and \: \beta \: are \: the \: \: roots \: of \: the \: equation \: = 7 {x}^{2} -3x + 2 = 0 \\ \\ \sf \to \: sum \: of \: roots \: of \: equation \: = \alpha + \beta = - \dfrac{b}{a} \\ \\ \sf \to \: \alpha + \beta = \dfrac{3}{7} \\ \\ \sf \to \: product \: of \: roots \: of \: equation \: = \alpha \beta = \dfrac{c}{a} \\ \\ \sf \to \: \alpha \beta = \dfrac{2}{7}$

$\sf \to \: value \: of \: = \dfrac{ \alpha }{1 - { \alpha }^{2} } + \dfrac{ \beta }{1 - { \beta }^{2} } \\ \\ \sf \to \: take \: lcm \: of \: the \: equation \\ \\ \sf \to \: \dfrac{ \alpha (1 - { \beta }^{2}) + \beta (1 - { \alpha }^{2} ) }{(1 - \alpha {}^{2} )(1 - { \beta }^{2}) } \\ \\ \sf \to \: \dfrac{ \alpha - \alpha { \beta }^{2} + \beta - \beta { \alpha }^{2} }{(1 - { \alpha }^{2})(1 - \beta {}^{2} )} \\ \\ \sf \to \: \dfrac{( \alpha + \beta ) - \alpha \beta ( \alpha + \beta )}{(1 - \alpha {}^{2})(1 - { \beta }^{2}) }$

$\sf \to \: \dfrac{( \alpha + \beta ) - \alpha \beta ( \alpha + \beta )}{1 - [( \alpha + \beta ) {}^{2} - 2 \alpha \beta] + ( \alpha \beta ) {}^{2} }$

=> 3/7 - 2/7X 3/7 / 1 - [ (3/7)² - 2 X 2/7] + 4/49

=> 15/49 / 1 + 19/49 + 4/49

=> 15/49 / 72/49

=> 15/49 X 49/72

=> 15/72 = 5 / 24

Hence option [ B ] is correct.