Question

solveeeeeeeeeeee 50 pts question in detaile.....

( Q.21. => 8. plz

Answer 1

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Given,

Quadratic equation = ax² + bx + c.

α and ß are its zeroes.

Now,

= a { ( α² / ß ) + ( ß² / α ) } + b { ( α / ß ) + ( ß / α ) }

= a { ( α³ + ß³ ) / αß } + b { ( α² + ß² ) / αß }

We know the relationship between zeroes and coefficient of quadratic equation,

Equation = ax² + bx + c

Here,

 Coefficient of x² = a

 Coefficient of x = b

 Constant term = c

Sum of zeroes = -b / a

 α + ß = -b / a.

Product of zeroes = c / a

 αß = c / a.

Now,

⇒ α² + ß² = ( α + ß )² - 2αß

⇒ α² + ß² = ( -b/a )² - 2 ( c/a )

⇒ α² + ß² = ( b² / a² ) - ( 2c / a )

⇒ α² + ß² = ( b² - 2ac ) / a²

Now,

⇒ α³ + ß³ = ( α + ß ) ( α² + ß² - αß )

⇒ α³ + ß³ = ( -b/a )  [ { ( b² - 2ac )/a² } - ( c/a ) ]

⇒ α³ + ß³ = ( -b/a ) { ( b² - 2ac - ac ) } / a²

⇒ α³ + ß³ = ( -b/a )  ( b² - 3abc ) ./ a²

⇒ α³ + ß³ = ( -b³ + 3abc ) / a³.

Now,

= a { ( α³ + ß³ ) / αß } + b { ( α² + ß² ) / αß }

= a { ( -b³ + 3abc ) / a³ ÷ ( c / a ) } + b { ( b² - 2ac ) / a² ÷ ( c /a ) }

= a [ { ( -b³ + 3abc ) / a³  } × ( a/c ) ] + b [ { ( b² - 2ac ) / a² } × ( a/c ) ]

= a  { ( -b³ + 3abc ) / a²c } + b { ( b² - 2ac ) / ac }

= { ( -b³ + 3abc ) / ac } + { ( b³ - 2abc ) / ac }

= ( -b³ + 3abc + b³ - 2abc ) / ac

= ( abc ) / ac

= b.

Hope it helps !!

Answer 2

Hey mate!

Kindly refer to the given attachments above!

Hope it helps you!