Question

If a,ß are the roots of x² – 3x + a=0 and
y,d are the roots of x2 - 12x + b = 0 and
a,B,X,8 in that order from a geometric
progression in increasing order with common
ratio r > 1, then a+b
​

Answer 1

Answer:

$a \: + \: b \: = \: 17 \: a$

Step-by-step explanation:

Given--->

$\alpha \: and \: \beta \: are \: the \:roots \: of \:$

${x}^{2} \: - \: 3x \: + \: a \: = \: 0$

$and \: \gamma \: and \: d \: are \: zeroes \: of \:$

${x}^{2} \: - 12 \: x \: + \: b \: = \: 0$

$and \: \alpha \: \beta \: \gamma \: and \: d \: are \: in \: geometric \: progression$

To find ---->

$value \: of \: ( \: a \: + \: b \: )$

Concept used ---->

1)

$if \: a {x}^{2} \: + \: bx \: + c \: = 0 \: is \: a \: quadratic \: equation \:$

$then$

$sum \: of \: zeroes \: = \: - \dfrac{b}{a}$

$product \: of \: zeroes \: = \: \dfrac{c}{a}$

2)

${ \: (a \: - \: b \: )}^{2} \: = \: ( \: a \: + \: b \: ) ^{2} \: - 4 \: a \: b$

3)

$if \: a \: b \: c \: and \: d \: are \: in \: gp \: then$

$\dfrac{b}{a} \: = \dfrac{d}{c}$

Solution----> ATQ,

${x}^{2} \: - 3 \: x \: + \: a \: = 0$

$\alpha \: + \: \beta \: = - \: \dfrac{ - 3}{1} = \: 3$

$\alpha \: \beta \: = \dfrac{b}{1} \: = \: b$

${x}^{2} \: - \: 12 \: x \: + \: b \: = \: 0$

$\gamma \: + \: d \: = - \dfrac{ - 12}{1} = 12$

$\gamma \: d \: = \dfrac{b}{1}$

$now \: \alpha \: \beta \: \gamma \: and \: d \: are \: in \: gp \: so \:$

$\dfrac{ \beta }{ \alpha } \: = \: \dfrac{d}{ \gamma }$

$= > \dfrac{ \alpha }{ \beta } \: = \dfrac{ \gamma }{d}$

$applying \: componendo \: and \: dividendo \: we \: get$

$\dfrac{ \alpha \: + \: \beta }{ \alpha \: - \: \beta } \: = \dfrac{ \gamma + \: d}{ \gamma \: - \: d}$

$squaring \: both \: sides \: we \: get$

$\dfrac{ { ( \: \alpha \: + \: \beta \: ) }^{2} }{( \: \alpha \: - \: \beta )^{2} } \: = \dfrac{( \: \gamma \: + d \: )^{2} }{ {( \: \gamma \: - \: d \: )}^{2} }$

$= > \dfrac{ {( \: \alpha \: + \: \beta \: )}^{2} }{( \: \alpha \: + \: \beta )^{2} \: - \: 4 \alpha \: \beta } \: = \dfrac{( \: \gamma \: + \: d \: )^{2} }{ (\gamma \: + \: d \: ) ^{2} \: - 4 \: \gamma \: d }$

$= > \dfrac{( \: 3 \:) ^{2} }{( \: 3 \: )^{2} \: - 4 \: a } \: = \dfrac{ {( \: 12 \: )}^{2} }{( \: 12 \: )^{2} \: - 4 \: b }$

$= > \dfrac{9}{9 \: - \: 4 \: a} \: = \dfrac{144}{144 \: - \: 4b}$

$= > \dfrac{1}{9 \: - \: 4a} \: = \dfrac{16}{144 \: - 4 \: b}$

$= > 144 \: - \: 4b \: = \: 16 \: ( \: 9 \: - \: 4a \: )$

$= > 144 \: - \: 4b \: = \: 144 \: - \: 64a$

$= > - 4b \: = \: - 64 \: a$

$= > 4b \: = \: 64a$

$= > b \: = \: 16 \: a$

$now$

$a \: + \: b \: = \: a \: + \: 16 \: a$

$a \: + \: b \: = 17 \: a$