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

Answer 1

If x = -2, the equation is satisfied
(-2)^2 + (a+3)(-2) = 2 b
4 - 2 a - 6 = 2 b
2 a + 2 b = -2
a + b = -1
so b = -a -1

If x = -3 the equation is satisfied
(-3)^2 +(a+3)(-3) = 2 b
9 - 3 a - 9 = 2 b
3 a + 2 b = 0

3 a + 2 (-a -1) = 0
3 a - 2 a = 2
a = 2
b = -3

Answer 2

another solution:
recall that a quadratic equation can also be rewritten as,
x^2 - (sum of roots)x + (product of roots) = 0
where
the numerical coeff of x = -(sum of roots), and
constant = (product of roots)

therefore,
sum of roots = -2 + (-3) = -5
equating this to the numerical coeff of x,
-(a + 3) = -5
a + 3 = 5
a = 2

product of roots = (-2)(-3) = 6
equating this to constant,
-2b = 6
b = -3

hope this helps~ :)