Question

please solve this question

Answer 1

▶ Question :- 

→ If α and β 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 
$\begin{lgathered}\frac{1}{ \alpha_1 \beta } + \frac{1}{ \alpha \beta_1 } + \frac{1}{ \alpha \alpha_1 } + \frac{1}{ \beta \beta_1 } . \\\end{lgathered}$


▶ Answer :- 


→ 1st Case :— 


α and β are the roots of the equation x² - px + q = 0 .

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


→ 2nd Case :—

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

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


▶ Now, 


→ Let's find :— 

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



✔✔ Hence, 1 is the answer ✅✅ .



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