Question

if alpha and bita are the zeroes of polynomial x^2 +6x+9 ,then form a polynomial whose zeroes are - alpha and - bita​

Answer 1

Step-by-step explanation:

:- x² + 6x + 9

x² + 3x + 3x + 9

x ( x + 3 ) + 3 ( x + 3 )

( x + 3 ) ( x + 3 )

• ( x + 3 ) = 0

x = ( - 3 )

• ( x + 3 ) = 0

x = ( - 3 )

• The new equation having Zeros as :-

So, the Zeros are 3

• Sum of the Zeros are :-

3 + 3 = 6

• Product of the Zeros are :-

3 × 3 = 9

♯ To form the quadratic equation we have formula as :-

x² - ( sum of Zeros )x + ( product of Zeros)

Putting value in it !!

x² - 6x + 9 is the required quadratic equation !!

Answer 2

Answer:

x²-6x-9

Step-by-step explanation:

we are given that alpha(a) and beta(b) are the zeroes of p(x)= x²+6x+9

therefore,

a+b = -6/1

ab= 9/1

now the zeroes of the other polynomial are -a and -b

-a+(-b)=-( a+b)= -(-6)/1= 6/1

and,

-a×(-b)= -(ab) = -9/1

on comparison

the required polynomial is

x²-6x-9

HOPE IT HELPS

PLEASE MARK BRAINLIEST