Signs of X^2 + p*X + q = 0

[Vlatko]

How can i find wheather the roots of the square equation (X^2 + p*X + q = 0) are positive or negative. (without solving the equation).

[fbokker]

the equation solves always like:

x = -p/2 +- Sqrt((p/2)^2-q)

so just check the expression in the root. If it is negative, you'll have to deal with imaginary numbers or in other words, there are no real solutions to your equation.

[Guv]

If the roots are complex, it is not clear what you mean by positive or negative.

For real roots, it helps to know that q is the product of the roots, and -p is their sum.

If q is negative, one root is positive and one is negative.

If q is positive, both roots have the same sign, which is the opposite of the sign of p.

[fbokker]

ok, let's try it again.
you never can say whether a squareroot is positive or negative. it is always both. think about
(-2)^2=2^2
sqareroot just undos the operation of ^2, but because the result of ^2 is always positive then information about the sign is lost. thus you usually have in quadratic equations the following two solutions:

x1 = -p/2 + Sqrt((p/2)^2-q)
x2 = -p/2 - Sqrt((p/2)^2-q)

let's use s= (p/2)^2-q, means the expression in the squareroot

in case of s=0 you will have one solution (x1=x2)
in case of s<0 there are only imaginary results (=no real solutions)
in case of s>0 you will always have two solutions which you have to check for sign

you can also write the quadratic equation as follows:
x^2+px+q=0=(x-x1)*(x-x2)
from this you get
q=x1*x2
-p=x1+x2
deducting further:
if q has a positive sign, both solutions x1 and x2 must have the same sign, else there will be one positive and one negative solution.
in case q is positive (x1 and x2 have the same sign) you can check the sign of them by taking a look at p: if p is positive, both have negative sign, else positive.
(which is somewhat the same as guv wrote, just with the mathematical background)

hope i didn't mix up the signs, i am close to braindead!
... and next time we'll show algebraic La Place transformations in visual basic (just kidding! but i could really use that)

[Vlatko]

OK, thanks for the replies.