Question

it is given that root 2 and -root 2 are 2 zeroes of the polynomial f(y)=2y4-3y3-3y2+6y-2 find all the zeroes of f(y)

Answer 1

Hey Mate !

Here is your solution :

Given,

f( y ) = 2y^4 - 3y^3 - 3y^2 + 6y - 2

☆ √2 and -√2 are its two zeroes.

Here,

=> Coefficient of y^4 ( a ) = 2

=> Coefficient of y^3 ( b ) = -3

=> Coefficient of y^2 ( c ) = -3

=> Coefficient of y ( d ) = 6

=> Constant term ( e ) = -2

Let, α and β be its other 2 zeroes.

We know the relationship between zeroes and coefficients of x.


=> Sum of zeroes = -b / a

=> α + β + √2 - √2 = - ( -3 ) / 2

•°• α + β = 3/ 2 ----------- ( 1 )

"And "

=> Product of zeroes = e/a

=> αβ × √2 × ( - √2 ) = -2 / 2

=> -2 αβ = -1

=> αβ = -1 ÷ ( -2 )

•°• αβ = ( 1/2 ) ------- ( 2 )

Now,

Using identity :

[ ( a - b )² = ( a + b )² - 4ab ]

=> ( α - β )² = ( α + β )² - 4αβ

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


=> ( α - β )² = ( 3/2 )² - 4 ( 1/2 )

=> ( α - β )² = ( 9/4 ) - 2

=> ( α - β )² = ( 9 - 8 ) / 4

=> ( α - β )² = 1/4

=> ( α - β ) = √( 1/4 )

•°• ( α - β ) = ( 1/2 ) ----------- ( 3 )

Adding ( 1 ) and ( 3 ),

=> a + β + α - β = ( 3/2 ) + ( 1/2 )

=> 2α = ( 3 + 1 ) / 2

=> 2α = 4/2

=> 2α = 2

=> α = 2 ÷ 2

•°• α = 1

Substituting the value of ( α ) in ( 2 ),

=> aβ = 1/2

=> 1 ( β ) = 1/2

•°• β = 1/2


Hence the other 2 zeroes are 1 and ( 1/2).

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Hope it helps !! ^_^