Question

if -6 and 3 are zeroes of the quadratic polynomial x^2 + (a + 1)x + b, then a and b are​

Answer 1

Answer:

2 and - 18

Step-by-step explanation:

Polynomials written in form of x^2 - Sx + P, represent S as sum of roots and P as product of roots.

So, here,

Sum of roots = - ( a + 1 )

Product of roots = b

As - 6 and 3 are roots of this polynomial,

Sum of roots = - 6 + 3

= > - ( a + 1 ) = - 6 + 3

= > - a - 1 = - 3

= > 3 - 1 = a

= > 2 = a

Product of roots = - 6*3

= > b = - 6*3 = - 18

Answer 2

Solution:-

The quadratic polynomial is x² + ( a + 1 )x + b

x = - 6 and 3

case - 1 :- x = - 6

put the value of x

In quadratic polynomial is x² + ( a + 1 )x + b = 0

We get

( - 6 )² + ( a + 1 ) × - 6 + b = 0

36 - 6a - 6 + b = 0

30 - 6a + b = 0

6a - b = 30 ......( i ) eq

Case - 2 :- x = 3

put the value of x

In quadratic polynomial is x² + ( a + 1 )x + b = 0

We get

( 3 )² + ( a + 1 ) × 3 + b = 0

9 + 3a + 3 + b = 0

3a + b + 12 = 0

3a + b = - 12 ........( ii ) eq

Add (i ) and (ii) eq

6a - b + 3a + b = 30 - 12

9a = 18

a = 2

Now put the value of a on (i) eq

6a - b = 30

6 × 2 - b = 30

12 - b = 30

- b = 30 - 12

- b = 18

b = - 18

Answer:- a = 2 and b = - 18