Question

If zeroes of the polynomial p(x)=x^2+(a+b)x+b are 2 and -3, then find the value of (a+b)

Answer 1

Answer:

p(x) = x²+(a+b)x+b

p = 1

q = (a+b)

r = b

Let α & β be the zeroes of p(x),

α = 2

β = -3

α+β = -q/p

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

-1 = -a+b

b = a-1 -----------(1)

αβ = r/p

(2)(-3) = b/1

-6 = b/1

-6 = b

b = -6

Substitute value of b in equation (1),

-6 = a-1

-6+1 = a

a = -5

So a = -5 and b = -6

a+b = -5+(-6) = -5-6 = -11