If m and n are roots of x^2 + mx + n = 0, and m and n are not 0, then the sum of the roots is...?

Answer 1

Let p(x) = x^2 + mx + n

If m is a root, then

p(m) = 0. But p(m) is also equal to plugging in the value m for x.
p(m) = m^2 + m(m) + n = m^2 + m^2 + n = 2m^2 + n

Therefore, 2m^2 + n = 0

If n is a root, then
p(n) = 0, and
p(n) = n^2 + mn + n

Therefore, n^2 + mn + n = 0

So we have a system of equations:
2m^2 + n = 0
n^2 + mn + n = 0

From the first equation, n = -2m^2, so substituting this into the second equation, we get

(-2m^2)^2 + m(-2m^2) + (-2m^2) = 0
4m^4 - 2m^3 - 2m^2 = 0

Since m is nonzero, we can divide both sides by 2m^2 to get

2m^2 - m - 1 = 0
(2m + 1) (m - 1) = 0
m = 1, m = -1/2

It follows that if m = 1, n = -2

m = -1/2, n = -1/2

In both cases, the sum of the roots is -1.

Answer 2

The last answers are misleading!
By the quadratic formula the roots are
(-m + √m²-4n)/2
and
(-m - √m²- 4n)/2
So the sum of the roots is -m, even if m or n is 0.
BTW The product of the roots is always n.
These results can often be handy in some
problems involving quadratics.

Answer 3

Then (x-m)(x-n) = x^2+mx+n, so -m = m+n, and mn = n. Therefore, m=1, and n = -2m, so n = -2, so m+n = -1.

x^2+x-2 = (x+2)(x-1).

Steve