Question

if alpha and beta r the roots of the equation x square - px+q=0 and alpha 1, beta 1 be the roots of the equation x square - px+p=o, then find 1 by alpha1beta+ 1by alphabeta1+1by alphaalpha1+1bybetabeta1

Answer 1

▶ Error :-

→ In equation 2 , it will be x² - qx + p = 0 instead of x² - px + p = o .


▶ Question :-

→ If $\alpha$ and $\beta$ are the roots of the equation x² - px + q = 0 and $\alpha _1$ , $\beta _1$ be the roots of the equation x² - qx + p = 0 , then find
$\frac{1}{ \alpha_1 \beta } + \frac{1}{ \alpha \beta_1 } + \frac{1}{ \alpha \alpha_1 } + \frac{1}{ \beta \beta_1 } .$


▶ Answer :-


→ 1st Case :—


$\alpha$ and $\beta$ are the roots of the equation x² - px + q = 0 .

Then,
$\sf \implies \alpha + \beta = \frac{ - b}{a} = \frac{ - ( - p)}{1} = p. \\ \\ and \\ \\ \sf \implies \alpha \beta = \frac{c}{a} = \frac{q}{1} = q.$


→ 2nd Case :—

$\alpha _1$ and $\beta _1$ are the roots of equation x² - qx + p = 0 .

Then,
$\sf \implies \alpha _1 + \beta _1 = \frac{ - b}{a} = \frac{ - ( - q)}{1} = q. \\ \\ and \\ \\ \sf \implies \alpha _1 \beta _1 = \frac{c}{a} = \frac{p}{1} = p.$


▶ Now,


→ Let's find :—

$\sf = \frac{1}{ \alpha_1 \beta } + \frac{1}{ \alpha \beta_1 } + \frac{1}{ \alpha \alpha_1 } + \frac{1}{ \beta \beta_1 } . \\ \\ \sf = \frac{1}{ \alpha_1 \beta } + \frac{1}{ \alpha \alpha_1 } + \frac{1}{ \alpha \beta _1 } + \frac{1}{ \beta \beta_1 } . \\ \\ \sf = \frac{1}{ \alpha_1 } ( \frac{1 }{ \alpha } + \frac{1}{ \beta } ) + \frac{1}{ \beta_1 } ( \frac{1}{ \alpha } + \frac{1}{ \beta } ). \\ \\ \sf = ( \frac{1 }{ \alpha _1} + \frac{1}{ \beta_1 } )( \frac{1}{ \alpha } + \frac{1}{ \beta } ). \\ \\ \sf = ( \frac{ \beta_1 + \alpha_1 }{ \alpha _1 \beta _1} )( \frac{ \beta + \alpha }{ \alpha \beta } ). \\ \\ \sf = \frac{ \cancel q}{ \cancel p} \times \frac{ \cancel{p}}{ \cancel q} . \\ \\ \huge \boxed{ \boxed{ \pink{ \sf = 1.}}}$



✔✔ Hence, 1 is the answer ✅✅ .



$\huge \boxed{ \green{ \mathcal{THANKS}}}$

Answer 2

$\alpha$ and $\beta$ are the roots of the equation x² - px + q = 0 .


$\implies \alpha + \beta = \frac{ - b}{a} = \frac{ - ( - p)}{1} = p. \\ \\ and \\ \\ \implies \alpha \beta = \frac{c}{a} = \frac{q}{1} = q.$




$\alpha _1$ and $\beta _1$ are the roots of equation x² - qx + p = 0 .


$\implies \alpha _1 + \beta _1 = \frac{ - b}{a} = \frac{ - ( - q)}{1} = q. \\ \\ and \\ \\ \implies \alpha _1 \beta _1 = \frac{c}{a} = \frac{p}{1} = p.$




$= \frac{1}{ \alpha_1 \beta } + \frac{1}{ \alpha \beta_1 } + \frac{1}{ \alpha \alpha_1 } + \frac{1}{ \beta \beta_1 } . \\ \\ = \frac{1}{ \alpha_1 \beta } + \frac{1}{ \alpha \alpha_1 } + \frac{1}{ \alpha \beta _1 } + \frac{1}{ \beta \beta_1 } . \\ \\ = \frac{1}{ \alpha_1 } ( \frac{1 }{ \alpha } + \frac{1}{ \beta } ) + \frac{1}{ \beta_1 } ( \frac{1}{ \alpha } + \frac{1}{ \beta } ). \\ \\ = ( \frac{1 }{ \alpha _1} + \frac{1}{ \beta_1 } )( \frac{1}{ \alpha } + \frac{1}{ \beta } ). \\ \\ = ( \frac{ \beta_1 + \alpha_1 }{ \alpha _1 \beta _1} )( \frac{ \beta + \alpha }{ \alpha \beta } ). \\ \\ = \frac{ q}{ p} \times \frac{ {p}}{ q} . \\ \\ = 1.$