If you have an equation of the form ax2 + c = 0, with a > 0, under what conditions will there be no real solutions?

If c>0, then the solution is imaginary, which is not a real solution.

Step-by-step explanation:

ax^2 + c = 0, with a > 0

Subtract c from each side

ax^2 +c-c = 0-c

ax^2 = -c

Divide each side by a

ax^2/a = -c/a

x^2 = -c/a

Take the square root of each side

sqrt(x^2) = sqrt(-c/a)

H

The only way to have real square roots is for -c/a to be positive.

We know that a>0, so -c >0

-c>0

Divide each side by -1, remembering to flip the inequality

c<0

If c<0  we have real solutions

If c=0  then x=0 which is a real solution

If c>0, then the solution is imaginary, which is not a real solution.

3-4

step-by-step explanation:

the answer is 3-4 because both 45 and 60 has a multiple of 15.

45/3=15

60/4=15

therefore the odds against selecting a red is 4-3