Question

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

Answer 1

sum of roots = -b/a
-3 + 2 = -(a+1)/1
-1 = -a -1
a =0
product of root = c/a
2× -3 = b/1
b = -6
so a+b = -6+0 = -6

Answer 2

sum pf zeros=α+β=2-3= -1
but sum =-b/a= -(a+1)/1=
-a-1=-1
-a=0
a=0
also product of zeros=αβ=2 x-3=-6
but product=c/a=b/1
therefore=-6
a=0...b=-6
........................plz mark brainliest.........................