The roots of the equation x^2+(a+3)x-2b=0 are -2 and -3.Find the values of a and b.
(x+2)(x+3) =x^2 + 5 x + 6
so
(a+3) = 5
-2b =6
a = 2
b = -3
check
x^2 + 5 x + 6 = 0
(x+2)(x+3)=0
so
(a+3) = 5
-2b =6
a = 2
b = -3
check
x^2 + 5 x + 6 = 0
(x+2)(x+3)=0