Question

please solve this question​

Answer 1

Question :-

If the zeroes of the quadratic polynomial x² + (a + 1)x + b are 2 and - 3 then find a and b.

Solution :-

Let f( x ) = x² + (a + 1)x + b

We know that the zeroes of the polynomial is the value the x of which polynomial becomes 0

Given :-

Zeroes of the given polynomial are 2 and - 3.

So, f( 2 ) = 0 and f( - 3 ) = 0

i ) f( 2 ) = 0

⇒ 2² + (a + 1)2 + b = 0

⇒ 4 + 2a + 2 + b = 0

⇒ 6 + 2a + b = 0

⇒ 2a + b = - 6 → eq(1)

ii ) f( - 3 ) = 0

⇒ ( - 3 )² + (a + 1)( -3 ) + b = 0

⇒ 9 - 3a - 3 + b = 0

⇒ - 3a + b + 6 = 0

⇒ 3a - b - 6 = 0

⇒ 3a - b = 6 → eq(1)

Adding eq(1) and eq(2)

⇒ 2a + b + (3a - b) = - 6 + 6

⇒ 2a + b + 3a - b = 0

⇒ 5a = 0

⇒ a = 0/5

⇒ a = 0

Substituting a = 0 in eq( 1 )

⇒ 2a + b = - 6

⇒ 2(0) + b = - 6

⇒ 0 + b = - 6

⇒ b = - 6

Hence, the value of a is 0 and the value of b is - 6.

Answer 2

Answer:

Value of a = 0

Value of b = -6

Step-by-step explanation:

Let f ( x ) = x² + ( a + 1 ) x + b

2 and - 3 are tge zeros of polynomials.

• (1) f ( 2 ) = 0
• (2) f ( - 3 = 0

(1) f ( 2 ) = 0

= 2² + ( a + 1 ) 2 + b = 0

= 4 + 2a + 2 + b = 0

= 6 + 2a + b = 0

= 2a + b = - 6 _[Equation]_(1)_

(2) f ( -3 ) = 0

= ( - 3 )² + ( a + 1 ) / ( - 3 ) + b = 0

= 9 - 3a - 3 + b = 0

= - 3a + b + 6 = 0

= 3a - b - 6 = 0

= 3a - b = 6 _[Equation]_(2)_

Adding: [Equation] (1)_and_(2)

= 2a + b + ( 3a - b ) = - 6 + 6

= 2a + b + 3a - b = 0

= 5a = 0

= a = 0 / 5

= a = 0

Adding: (a = 0) in [Equation]_(1)_

= 2a + b = - 6

= 2 ( 0 ) + b = - 6

= 0 + b = - 6

= b = - 6

Therefore , value of a = 0 and value of b = -6