Question

Find the equation whose roots are:-3,2/3,1/2​

Answer 1

Answer:

Here, S

1

=0+1+2+(−3)=0,

S

2

=0.1+1.2+2.(−3)+(−3).0+1.(−3)+0.2=−7

and S

3

=0.1.2+1.2.(−3)+2.(−3).0=−6,S

4

=0.1.2.(−3)=0

Thus required bi-quadratic is, x

4

−S

1

x

3

+S

2

x

2

−S

3

x+S

4

=0⇒x

4

−7x

2

+6x=0

Answer 2

Answer:

6x³ + 11x² - 19x + 6 = 0

Step-by-step explanation:

Given :- The given roots are - 3, 2/3 and 1/2.

To Find :- The equation whose roots are given.

Solution :-

The given roots are - 3 , 2/3 , 1/2 .

Since there are three roots, therefore it is a cubic equation.

A cubic polynomial equation is an equation whose variable has a highest degree of 3.

The cubic equation will be :-

( x - ( - 3 ) )( x - ( 2/3 ) )( x - ( 1/2 ) ) = 0

⇒ ( x + 3 )( ( 3x - 2 )/3 )( ( 2x - 1 )/2 ) = 0

⇒ (x + 3)( 3x - 2 )( 2x - 1 ) = 0

⇒ ( 3x² - 2x + 9x - 6 )( 2x - 1 ) = 0

⇒ 6x³ - 4x² + 18x² - 12x - 3x² + 2x - 9x + 6 = 0

⇒ 6x³ + 11x² - 19x + 6 = 0

Therefore, the equation whose roots are - 3, 2/3 and 1/2 is 6x³ + 11x² - 19x + 6 = 0

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