Irrational numbers can not be repeated or terminate. So the answer is IRRATIONAL
You have shared the situation (problem), except for the directions: What are you supposed to do here? I can only make a educated guesses. See below:
Note that if <span>ax^2+bx+5=0 then it appears that c = 5 (a rational number).
Note that for simplicity's sake, we need to assume that the "two distinct zeros" are real numbers, not imaginary or complex numbers. If this is the case, then the discriminant, b^2 - 4(a)(c), must be positive. Since c = 5,
b^2 - 4(a)(5) > 0, or b^2 - 20a > 0.
Note that if the quadratic has two distinct zeros, which we'll call "d" and "e," then
(x-d) and (x-e) are factors of ax^2 + bx + 5 = 0, and that because of this fact,
- b plus sqrt( b^2 - 20a )
d = ------------------------------------
2a
and
</span> - b minus sqrt( b^2 - 20a )
e = ------------------------------------
2a
Some (or perhaps all) of these facts may help us find the values of "a" and "b." Before going into that, however, I'm asking you to share the rest of the problem statement. What, specificallyi, were you asked to do here?
Answer:
1 I don't know if those are questions but try if they are try the first one a
Let the number of baskets be x
Let the number of points be y
32 +3x = y
19 + 4x = y
Rearrange both equations then solve them as simultaneous equations
32 = y - 3x
19 = y - 4x
y - 3x = 32
- y - 4x = 19
x = 13
Substitute x into one of the equations
y - 3x = 32
y - (3 x 13) = 32
y - 39 = 32
y = 32 + 39
y = 71
Number of baskets (x) = 13
Number of points (y) = 71