Question

answer the question in the given picture​

Answer 1

Correct option is

B

0

(ax+b)(3x+2)=0

⇒ 3ax

2

+2ax+3bx+2b

⇒ 3ax

2

+(2a+3b)x+2b=0

It is given that

3

−2

and

2

1

are zeros of the equation 3ax

2

+(2a+3b)x+2b=0

Here, A=3a,B=2a+3b,C=2b

Let α=

3

−2

and β=

2

1

We know,

⇒ α+β=

A

−B

⇒

3

−2

+

2

1

=

3a

−(2a+3b)

⇒

6

−4+3

×3a=−2a−3b

⇒

2

−a

=−2a−3b

⇒ −a=−4a−6b

⇒ 3a+6b=0 ----- ( 1 )

Now,

⇒ α.β=

A

CA

⇒

3

−2

×

2

1

=

3a

2b

⇒

3

−1

×3a=2b

⇒ −a+2b=0 ----- ( 2 )

Adding equation ( 1 ) and ( 2 ) we get,

b=0

Put b=0 in ( 2 ) we get,

a=0

∴ a+b=0+0=0