Math, asked by MichhDramebaz, 5 days ago

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❥ If α and β are the roots of the equation ax² + bx + c =0, express the roots of the equation a³x² - ab²x + b²c = 0 in terms of α and β.

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Answers

Answered by TheHoneyBabe
6

Step-by-step explanation:

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Answered by XxitsmrseenuxX
10

Answer:

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 \large\underline {\sf  {\pink{ \pink\star {\: GIVEN:  - }}}} \\

 \sf{ \rightarrow \: If  \: a  \: and  \: B are \:  the \:  roots  \: of  \: the  \: equation \: } \\  \sf{   ax  ^{2}  + bx + c =0,}</p><p> \\  \\

 \large\underline {\sf  {\red{ \pink\star {\: TO \: FIND :  - }}}} \\

 \sf{ \rightarrow \: express  \: the  \: roots \:  of \:  the  \: equation  \:} \\  \sf{  {a}^{3} {x}^{2}   - ab ^{2} x + b ^{2} c = 0  \: in  \: terms  \: of  \: }  \\  \sf{a \:  and \:  B \: .}</p><p> \\  \\

 \large\underline {\sf  {\orange{ \pink\star {\:  SOLUTION \: :  - }}}} \\

 \sf\green{   \alpha  \:  and  \beta \:  are  \: the  \: roots  \: of  \: the \:  equation. \: } \\

\sf{ \Rightarrow \:  ax ^{2} + bx + c = 0.} \\  \\

 \sf \blue{As \:  we  \: know \:  that,} \\ \\

 \sf\green{Sum  \: of  \: the \:  zeroes \:  of \:  the \:  quadratic } \\  \sf{ \: polynomial.} \\ </p><p> \sf{   \Rightarrow\alpha  +  \beta  =  \frac{ - b}{a} } \\  \\

 \sf\green{Products \:  of  \: the \:  zeroes \:  of \:  the  \: quadratic  \: } \\\sf\green{ polynomial.} \\  \sf{  \Rightarrow \: \alpha  \beta  =  \frac{c}{a} } \\  \\

 \sf\green{Roots  \: of \:  the  \: equation} \\

\sf{ \Rightarrow \: a ^{3} \:  x ^{2} ab ^{2} x + b^{2} c = 0.} \\ \\ \\

 \sf \blue{As \:  we \:  know \:  that,} \\

\sf\green{Let, \:y \:and\: \delta \:are \:the \:roots\: of \:the \:equation.}\\

\sf{\Rightarrow a ^{3} \:  x ^{2} - ab ^{2} x + b^{2}c = 0.}\\ \\ \\

\sf\green{Sum \:of\: the \:zeroes \:of \:the \:quadratic\: }\\

\sf\green{polynomial.}\\

\sf{\Rightarrow \:y + \delta =  \frac{-b}{a}}\\

\sf{\Rightarrow \:y + \delta=\frac{(-ab^{2})}{a^{3}}  = ab^{2}/a^{3}.}\\

\sf{\Rightarrow \:y +\delta =  =(\: \frac{b^{2}}{a^{2}}\: )=(\: \frac{-b}{a}^{2}\: )= (\: \alpha +\beta \: )^{2}.}\\ \\ \\

\sf\green{Products \:of\: the \:zeroes \:of \:the \:quadratic} \\

\sf\green{polynomial.}\\

\sf{\Rightarrow \: y \delta = \frac{c}{a}}\\

\sf{\Rightarrow \: y \delta =(\frac{b^{2}c}{a^{3}})=(\frac{b^{2}}{a^{2}}) (\frac{c}{a}) }\\

\sf{\Rightarrow \:y \delta =(\: \frac{b}{a}^{2}\: )(\frac{c}{a})= (\: \alpha +\beta \: )^{2}(\: \alpha +\beta \: ) }\\ \\ \\

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