Math, asked by mohdadnan1313, 10 months ago

form a quadratic polynomial whose zeroes 1/α & 1/β given that α and β are the zeroes of the polynomial 6x^2-7y+2

Answers

Answered by Anonymous
3

Answer:

2x² - 7x + 6

Step-by-step explanation:

Quadratic polynomial : 6x² - 7x + 2

α, β are the zeroes of the polynomial

Comparing 6x² - 7x + 2 with ax² + bx + c we get,

  • a = 6
  • b = - 7
  • c = 2

Sum of zeroes = α + β = - b / a = - ( - 7 ) / 6 = 7 / 6

Product of zeroes = αβ = c / a = 2 / 6

If 1 / α , 1 / β are zeroes of the polynomial

Sum of zeroes = 1 / α + 1 / β = ( α + β ) / αβ = ( 7 / 6 ) / ( 2 / 6 ) = 7 / 2

Product of zeroes = ( 1 / α ) × ( 1 / β ) = 1 / αβ = 1 / ( 2 / 6 ) = 6 / 2

Quadractic polynomial :

= k{ x² - ( Sum )x + Product }

[ Where k ≠ 0 ]

= k{ x² - ( 7/2 )x + 6/2 }

= k( x² - 7x/2 + 6/2 )

When k = 2

= 2( x² - 7x/2 + 6/2 )

= 2x² - 7x + 6

Hence the required quadratic polynomial is 2x² - 7x + 6.

Answered by madhav8656
0

Answer:

Step-by-step explanation:

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