Question

if one of the zeros of the quadratic polynomial x^2 +(a+1)x + b are 2,-3 find a and b.

Answer 1

Let the quadratic polynomial be:

p(x) = x² + (a + 1)x + b

The two zeros are 2 and -3.

So p(2) = 0
∴2² + (a+1)*2 + b = 0
∴4 + 2a + 2 + b = 0
∴2a + b = -6 -------------------- (1)

Also, p(-3) = 0
∴(-3)² + (a+1) * (-3)  + b = 0
∴9 -3a - 3 + b = 0
∴6 = 3a - b
∴3a - b = 6 ----------------- (2)

Solving the two equations by Elimination Method (You can use other methods also),

2a + b = -6
3a - b = 6                       (Adding (1) and (2)  )
----------------
∴ 5a = 0
∴ a = 0

Substituting a = 0 in equation (1),

2a + b = -6
∴2(0) + b = -6
∴0 + b = -6
∴b = -6

Thus a = 0 and  b = -6.

Answer 2

Given
Quadratic polynomial is x² + (a+1)x + b
zeroes of the polynomial are 2 and -3
sum of the zeroes = $\frac{a+1}{1}$ = 2-3
a+1 = -1
a = -2
product of the zeroes = $\frac{b}{1}$ = 2(-3)
b = -6