Question

if alfa, bita are the zeroes of polynomial y sq.-a(y+1)-b such that (alfa+1) (bita+1)=0 than b= ?​

Answer 1

Step-by-step explanation:

f(x)=x

2

+x+1

a=1

b=1

c=1

∵α and β are the zeroes of above polynomial.

∴ Sum of roots =

a

−b

⇒α+β=

1

−1

⇒α+β=−1⟶(1)

Product of roots =

a

c

⇒αβ=

1

1

⇒αβ=1⟶(2)

∴

α

1

+

β

1

=

αβ

β+α

From eq

n

(1)&(2), we have

⇒

α

1

+

β

1

=

1

−1

=−1

Hence, -1 is the correct answer.

Answer 2

Answer:

b=1

Step-by-step explanation:

let f(y)=y²-a(y+1)-b

=y²-ay-(a+b)

α,β are zeros ofd f(y)

so α+β=-(-a/1)=a

αβ=-(a+b)/1=-(a+b)

Now( α+1)(β+1)=0

=> αβ +(α+β)+1=0

=> -(a+b)+a+1=0

-a-b+a+1=0

b=1