Question

23 both parts dear geniuses??

Answer 1

ANSWER


EQUATION

${x}^{2} - 3x - 2 = 0$



$\alpha \: \: \beta \: are \: roots \:$



$= > \alpha . \beta = \frac{c}{a}$



$= \alpha . \beta = { - 2}$



And



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



$\alpha + \beta = \frac{ - 3 \times - 1}{1}$



$\alpha + \beta = 3$




FOR FINDING THE EQ WHOSE ROOTS ARE



$\frac{1}{2 \alpha + \beta } \: and \frac{1}{2 \beta + \alpha }$




NOTE



IF ROOTS ARE GIVEN AND WE HAVE TO FIND
EQ LET A AND B ARE ROOTS THEN EQ BECOME



${x}^{2} - (a + b) + (a \times b) = 0$
EQ.............(1)

so





WE HAVE FIND

$\frac{1}{2 \alpha + \beta } + \frac{1}{2 \beta + \alpha }$



$\frac{ \alpha + 2 \beta + 2 \alpha + \beta }{(2 \alpha + \beta )( \alpha + 2 \beta )}$



$= \frac{3( \alpha + \beta )}{ { \alpha }^{2} + 4 \alpha \beta + 2 { \beta }^{2} + \alpha \beta }$



$= \frac{3( \alpha + \beta )}{2 { \: ( \alpha + \beta) }^{2} + \alpha \beta }$



put value of
$= \alpha \: and \: \beta$


$= \frac{3 \times 3}{2 {(3)}^{2} - 2}$



$= \frac{9}{16}$

similarly find



$\frac{1}{2 \alpha + \beta } \times \frac{1}{2 \beta + \alpha }$



$= \frac{1}{16}$


put these value in eq ...(1)


we get
${x}^{2} - (a + b) + (a \times b) = 0$

so

${x }^{2} - \frac{9}{16} x + \frac{1}{16} = 0$


$16 { x }^{2} - 9x + 1 = 0$

HENCE

EQ IS

$16 { x }^{2} - 9x + 1 = 0$