Question

which equations have at least one positive solution for x:

-x^2+6x+2=0
-x^2+6x-8=0
1/4x^2+3x+8=0
1/4x^2+3x+12=0

Answer 1

-x^2 + 6x + 2 = 0 is equivalent to x^2 - 6x - 2 = 0. The discriminant is positive, so there will be two Real answers. They'll be ugly, but they exist, so we can factor the equation as (x + A)(x + B) where AB = -2, so one of them must be postive and one must be negative. The values of x are -A and -B, so whichever one is negative will yield a positive value of x. Therefore there must be one positive root.

I'll post this and then move on to the other three equations, so that the post isn't overly long.

Answer 2

-x^2 + 6x - 8 = 0 is equivalent to x^2 - 6x + 8 = 0. Again, this has 2 Real solutions. However, when we factor this as (x + A)(x + B) we find that AB = 8 meaning either A and B are both positive (giving us 2 negative values of x) or both negative (giving us 2 positive values of x). Since A + B must add to -6, we can conclude that both A and B are negative, but this gives us 2 positive values for x (because x = -A and x = -B).

Answer 3

(1/4)x^2 + 3x + 8 = 0 is equivalent to x^2 + 12x + 32 = 0. This has 2 real solutions.
Factoring it as (x + A)(x + B) we have AB = 32, A + B = 12. Since AB is positive, A and B must have the same sign. Since A + B = 12, which is positive, A and B must both be positive. Since the solutions are x = -A and x = -B, both values of x are negative. This equation does not have a positive solution.

Answer 4

(1/4)x^2 + 3x + 12 = 0 is equivalent to x^2 + 12x + 48 = 0.
The discriminant for this is 12^2 - 4(1)(48) = 144 - 192 = -48 which is negative, thus this quadratic equation has no real solutions and only real numbers can be positive, so it has no positive solutions.