Question

if zeros of the puadratic polynomial x2+(a+1)x+b are 2 and -3 find the value of a and b​

Answer 1

Answer:

$\large\boxed{\sf{a=0,\;\;b=-6}}$

Step-by-step explanation:

Given a quadratic polynomial such that,

${x}^{2}+(a+1)x+b$

Also, the roots of the polynomial are 2 and -3.

Now, we know that, General form of a quadratic polynomial is ,

$A{x}^{2}+Bx+C$

On Comparing the coefficients, we have,

• A = 1
• B = a + 1
• C = b

Now, we know that,

Sum of roots = -B/A

=> -(a+1)/1 = 2 +(-3)

=> -(a+1) = 2 -3

=> a + 1 = -(-1)

=> a = 1-1

=> a = 0

And, also, we know that,

Product of roots = C/A

=> b/1 = 2(-3)

=> b = -6

Hence, the values of a = 0 and b = -6.

Answer 2

Given :

Zeroes of the polynomial x² + ( a + 1 )x + b are 2 and - 3

To find :

Values of a and b

Solution :

Let p( x) = x² + ( a + 1 )x + b

2 is the zero of the given polynomial

We know that

The value of x for which the polynomial becomes 0

=> p( 2 ) = 0

=> 2² + ( a + 1 ) × 2 + b = 0

=> 4 + 2a + 2 + b = 0

=> 2a + b + 6 = 0 --- EQ( 1 )

- 3 is a zero of the given polynomial

=> p( - 3 ) = 0

=> ( - 3 )² + (a + 1)( - 3 ) + b = 0

=> 9 - 3a - 3 + b = 0

=> - 3a + b + 6 = 0 --- EQ( 2 )

From EQ( 1 ) & ( 2 )

=> 2a + b + 6 = - 3a + b + 6

=> 2a = - 3a

=> 2a + 3a = 0

=> 5a = 0

=> a = 0

Substituting a = 0 in Eq( 1 )

=> 2a + b + 6 = 0

=> 2( 0 ) + b + 6 = 0

=> b = - 6

Therefore the value of a is 0 and b is - 6.