Question

if one is the zero of 3x3-x2-3x+1, the the other two zeroes are -1 and 1/3. justify

Answer 1

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Answer 2

Let α , β and Г be its zeroes.

Given , 1 is one of its zeroes.

Let, Г = 1

P(x) = 3x³ - x² - 3 x + 1

Here,

Coefficient of x³ = 3

Coefficient of x² = -1

Coefficient of x = -3

Constant term = 1.

We have,

⇒ Sum of zeroes = - ( coefficient of x² ) ÷ ( coefficient of x³ )


⇒ α + β + Г = - ( -1 ) ÷ 3

⇒ α + β + 1 = 1/3

⇒ α + β = ( 1/3 ) - 1

⇒ α + β = -2/3. ------ ( 1 )

Now,

⇒ Product of zeroes = - ( constant term ) ÷ coefficient of x³

⇒ αβГ = -1/3

⇒ αβ × 1 = -1/3

∴ αβ = -1/3.

Now,

⇒ ( α - β )² = ( α + β )² - 4αβ

By substituting the values of ( α + β ) and αβ.

⇒ ( α - β )² = ( -2/3 )² - 4 ( -1/3 )

⇒ ( α - β )² = ( 4/9 ) + ( 4/3 )

⇒ ( α - β )² =  16/9

⇒ ( α - β ) = √16/9

∴  ( α - β ) = 4/3.  ------ ( 2 )

By adding ( 1 ) and ( 2 ),

⇒ α + β + α - β =( -2/3 ) + ( 4/3 )

⇒ 2α = 2/3

⇒ α =( 2/3 ) ÷ 2

∴  α = 1/3.

By substituting the value of α in ( 1 ),

⇒ α + β = -2/3

⇒ ( 1/3 ) + β = -2/3

⇒ β = ( -2/3 ) - ( 1/3 )

∴   β = -1.

Other 2 zeroes = 1/3 and ( -1 ).

Verification :

⇒ Sum of zeroes = - Coefficient of x² ÷ Coefficient of x³

⇒ 1 - 1 + 1/3 = - ( -1 ) / 3

∴ 1/3 = 1/3.

Now,

⇒ Sum of zeroes taken two at a time = Coefficient of x ÷ Coefficient of x³



⇒ { 1 ( -1 ) + (-1 ) ( 1/3 ) + 1 ( 1/3 ) } = -3 ÷ 3

⇒ { -1 - ( 1/3 ) + ( 1/3 ) } = -1

∴ -1 = -1

Verified.

Again,

⇒ Product of zeroes = - Constant term ÷ Coefficient of x³

⇒ 1 ( -1 ) ( 1/3 ) = -1/3

⇒ -1/3 = -1/3

Verified.