Question

if 2and-3 are the zeroes of x²+(a+1)x+b then find the values of a and b​

Answer 1

Given :

Given that, 2 and -3 are the zeroes of the polynomial x²+(a+1)x+b.

To find :

We should the values of a and b.

Solution :

Here, the given polynomial x²+(a+1)x+b is in the form of a quadratic equation where :

• a = 1
• b = (a+1)
• c = b

Sum of the zeroes = -b/a

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

Product of zeroes = c/a

• b/1 = 2 × (-3)
• b/1 = -6
• b = -6

Verification :

Let's substituting the values along with the zeroes and verify :

• x²+(a+1)x+b = 0
• (2)²+(0+1)(2)+(-6) = 0
• 4+(1)(2)-6 = 0
• 4+2-6 = 0
• 6-6 = 0
• LHS = RHS

Henceforth, verified!

Answer 2

• put 2 in p(x) = x² + (a+1)x + b
• p(2) = (2)² + (a+1)(2) + b = 0
• 4 + 2a + 2 + b = 0
• 2a + b + 6 = 0
• 2a = -(b + 6)
• a = -(b+6)/2 → eq(1)
• then ,
• put -3 in p(x) = x² + (a+1)x + b
• then ,
• (-3)² + (a+1)(-3) + b = 0
• 9 + -3a -3 + b = 0
• b - 3a = -9 + 3
• b = 3a + 6 = 3(-(b+6)/2) + 6 = (-3b-18+12)/2
• 2b = -3b - 6
• 2b + 3b = -6
• 5b = -6
• b = -⅚
• now put b = -⅚
• a = -(-⅚+6)/2 = (41/6)/2 = 41/6 × 2 = 41/3

Mark as Brainliest answer please my friend