Question

quadratic polynomial 2x2 -3 X + 1 has zero as alpha and beta . Now, form a quadratic polynomial whose zeros are 3 alpha and 3 Beta.

Answer 1

Heya !!!


P(X) = 2X² - 3X + 1



Here,



A = Coefficient of X² = 2



B = Coefficient of X = -3


And,


C = Constant term = 1




Sum of zeroes = - B/A



Alpha + Beta = -(-3)/2



Alpha + Beta = 3/2 -------(1)


And,


Product of zeroes = C/A



Alpha × Beta = 1/2 ------(2)





Sum of zeroes of quadratic polynomial whose zeroes are 3Alpha and 3 beta .



Sum of zeroes = 3Alpha + 3Beta




=> 3 ( Alpha + Beta )


=> 3 × 3/2


=> 9/2



And,



Product of zeroes = 3Alpha × 3Beta



=> 3 ( Alpha × Beta )



=> 3 × 1/2


=> 3/2



Therefore,


Required quadratic polynomial = X²-(Sum of zeroes)X + Product of zeroes



=> X² - ( 9/2)X + 3/2


=> X² - 9X /2 + 3/2


=> 2X² - 9X + 3



★ HOPE IT WILL HELP YOU ★

Answer 2

Here is your solution :

Given,

Quadratic equation = 2x² - 3x + 1

Here ,

Coefficient of x² ( a ) = 2

Coefficient of x ( b ) = ( -3 )

Constant term ( c ) = 1


( Alpha ) and ( Beta ) are its zeroes.

We know the relationship between zeroes and coefficient of x of quadratic equation.

=> Sum of zeroes = -b/a

=> Alpha + Beta = -( -3 ) / 2

•°• Alpha + Beta = 3/2 ------- ( 1 )

=> Product of zeroes = c/a

=> Alpha × Beta = ( 1/2 ) -------- ( 2 )


The formula for a quadratic equation is :

Zeroes : 3 Alpha and 3 Beta

=> x² - ( Sum of zeroes )x + Product of zeroes = 0


=> x² - ( 3 Alpha + 3 Beta )x + 3 Alpha × 3 Beta = 0

=> x² - 3( Alpha + Beta )x + 9 Alpha × Beta = 0

Substitute the value of ( 1 ) and ( 2 ),

=> x² - 3( 3/2 )x + 9( 1/2 ) = 0

=> x² - ( 9/2 )x + ( 9/2 ) = 0

=> ( 2x² - 9x + 9 ) / 2 = 0

•°• 2x² - 9x + 9 = 0

The required quadratic equation is ( 2x² - 9x + 9 ).