Is there more than one answer to the problem. if so can I see them.
Answer:
Yes
Step-by-step explanation:
Yes it is possible to solve a quadratic equation that is not factorable over the set of integers.
The solution may vary like Integers, rationals, irrationals or complex solutions.
To find two roots of the equation we can always use the formula given below to solve a quadratic equation,
For the quadratic equation,
, we have,

If the discriminant is greater than
, we get complex roots.
9514 1404 393
Answer:
y = 1 · 5^x
Step-by-step explanation:
Use the given values in the given formula and solve for 'a' and 'b'.
y = a·b^x
5 = a·b^1 . . . . . . for (x, y) = (1, 5)
125 = a·b^3 . . . . for (x, y) = (3, 125)
Divide the second equation by the first:
125/5 = (a·b^3)/(a·b^1)
25 = b^2 . . . simplify
5 = b . . . . . . . take the square root
Substituting into the first equation, we can find 'a':
5 = a·5
1 = a . . . . . . . divide by 5
Written in the given form, the equation is ...
y = 1 · 5^x
49,000 is the nearest round to the hundredth to 48,950
Answer:

Step-by-step explanation:
