Question

Form the quadratic polynomial whose zeroes are 1by3 and 1by3

Answer 1

AnsWer :

9x² - 6x + 1.

Solution :

We have,

Let Zeros be

• a = 1/3.
• b = 1/3.

A/Q,

K ( x² - Sx + P )

• K ( Contant term )
• S ( Sum of zeros )
• P ( Product of zeros )

For Sum of zeros.

=> a + b = 1/3 + 1/3

=> a + b = 2/3.

For Product of zeros.

=> ab = 1/3 * 1/3.

=> ab = 1/9.

Putting value,We get.

=> K ( x² - ( 2/3 )x + 1/9

=> k ( 9x² - 6x + 1 )

$\rule{200}3$

Let's Verify :

We have, Equation.

=> 9x² - 6x + 1 = 0.

=> 9x² - 3x - 3x + 1 = 0.

=> 3x( 3x - 1 ) - 1( 3x - 1 ) = 0.

=> ( 3x - 1 )( 3x - 1 ) = 0.

Either,

=> x = 1/3 Or, x = 1/3.

Answer 2

Step-by-step explanation:

AnsWer :

9x² - 6x + 1.

Solution :

We have,

Let Zeros be

a = 1/3.

b = 1/3.

A/Q,

K ( x² - Sx + P )

K ( Contant term )

S ( Sum of zeros )

P ( Product of zeros )

For Sum of zeros.

=> a + b = 1/3 + 1/3

=> a + b = 2/3.

For Product of zeros.

=> ab = 1/3 * 1/3.

=> ab = 1/9.

Putting value,We get.

=> K ( x² - ( 2/3 )x + 1/9

=> k ( 9x² - 6x + 1 )

Let's Verify :

We have, Equation.

=> 9x² - 6x + 1 = 0.

=> 9x² - 3x - 3x + 1 = 0.

=> 3x( 3x - 1 ) - 1( 3x - 1 ) = 0.

=> ( 3x - 1 )( 3x - 1 ) = 0.

Either,

=> x = 1/3 Or, x = 1/3.