Question

If alpha and beta are the zeros of the polynomial f(x) = x^2 - 3x - 2, find a quadratic polynomial whose zeros are 2 alpha / beta and 2 beta / alpha.

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Answer 1

Answer:

=> a + ß = - coefficient of x / coefficient of x^2

=> a + ß = - ( - 3 / 1 )

=> a + ß = - ( - 3 )

=> a + ß = 3

=> aß = - 2 / 1

=> aß = - 2

=> Sum of zero = 1 / 2 a + ß + 1 / 2 ß + a

=> 1 / 2 a + ß × 2 ß + a / 2 ß + a + 1 / 2 ß + a × 2 / 2

=> 2 ß + a / ( 2 a + ß ) ( 2 ß + a ) + 2 a + ß / ( 2 ß + a ) ( 2 a + ß )

=> 2 ß + a + 2 a + ß / ( 2 a + ß ) ( 2 ß + a ) = 3 ß + 3 a / 4 a ß + 2 ß^2 + 2 a^2 + ß a

=> 3 ( ß + a ) / 5 a ß + 2 ( a^2 + ß^2 )

=> 3 ( a + ß ) / 5 a ß + 2 [ ( a + ß )^2 - 2 a ß ]

=> 3 ( 3 ) / 5 ( - 2 ) + 2 [ ( 3 )^2 - 2 × ( - 2 ) ]

=> 9 / - 10 + 2 ( 13 )

=> 9 / - 10 + 26

=> Therfore, Sum of zeros = 9 / 16.

=> Product of zeros = 1 / 2 a + ß × 1 / 2 ß + a

=> 1 / ( 2 a + ß ) ( 2 ß + a )

=> 1 / 4 a ß + 2 ß^2 + 2 a^2 + ß a

=> 1 / 5 a ß + 2 ( a^2 + ß^2 )

=> 1 / 5 a ß + 2 [ ( a + ß )^2 - 2 a ß ]

=> 1 / 5 ( - 2 ) + 2 [ ( 3 )^2 - 2 × ( - 2 ) ]

=> 1 / 10 + 2 ( 9 + 4 )

=> 1 / 10 + 2 ( 13 )

=> 1 / - 10 + 26

=> Therefore, Product of zeros = 1 / 16.

=> f (x) = k ( x^2 - sx - p )

=> = x^2 - 9 / 16 x + 1 / 16.

Step-by-step explanation:

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