Problem 005 | Quadratic Equations

Submitted by Anonymous (not verified) on Wed, 02/20/2013 - 00:05

উদাহরণ ৫৷ [tex]{x^2} + 3x + 4 = 0[/tex] এই সমীকরণের বীজ দুটি [tex]\alpha ,\beta [/tex] হলে, যে সমীকরণের বীজ দুটি [tex]{\left( {\alpha  + \beta } \right)^2},{\left( {\alpha  - \beta } \right)^2}[/tex] তা নির্ণয় করো।                              [H.S ‘84]

সমাধান: [tex]{x^2} + 3x + 4 = 0[/tex] এই সমীকরণের বীজ দুটি হল [tex]\alpha ,\beta [/tex]।

অতএব  [tex]\alpha  + \beta  =  - 3 \to \left( 1 \right),\alpha \beta  = 4 \to \left( 2 \right)[/tex]

সুতরাং  [tex]{\left( {\alpha  + \beta } \right)^2} = {\left( { - 3} \right)^2} = 9 \to \left( 3 \right)[/tex]

আমরা জানি

[tex]\begin{array}{l}
{\left( {\alpha  - \beta } \right)^2}\\
 = {\left( {\alpha  + \beta } \right)^2} - 4\alpha \beta \left[ {by\left( 1 \right),\left( 2 \right)} \right]\\
 = {\left( { - 3} \right)^2} - 4 \times 4\\
 = 9 - 16\\
 =  - 7\\
 \Rightarrow {\left( {\alpha  - \beta } \right)^2} =  - 7 \to \left( 4 \right)
\end{array}[/tex]

(3),(4) ব্যবহার করে পাই

[tex]\begin{array}{l}
{\left( {\alpha  + \beta } \right)^2} + {\left( {\alpha  - \beta } \right)^2} = 9 - 7 = 2 \to \left( 5 \right)\\
{\left( {\alpha  + \beta } \right)^2} \times {\left( {\alpha  - \beta } \right)^2} = 9 \times \left( { - 7} \right) =  - 63 \to \left( 6 \right)
\end{array}[/tex]

(5) ও (6) ব্যবহার করে পাই

[tex]\begin{array}{l}
{x^2} - \left\{ {{{\left( {\alpha  + \beta } \right)}^2} + {{\left( {\alpha  - \beta } \right)}^2}} \right\}x + {\left( {\alpha  + \beta } \right)^2}{\left( {\alpha  - \beta } \right)^2} = 0\\
 \Rightarrow {x^2} - 2x - 63 = 0
\end{array}[/tex]

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