What can you say about the value of "a" if the equation y=ax^2-8 has no solutions?

a is not 4

Ok, thank you very much. I really appreciate it. Would you mind explaining how you came to this conclusion please?

a is an imaginary number. The only way this doesn't have an answer is if you use something like the square root of -5. Technically, there is an answer even for that, but that's beyond the scope of the equation.


dkgo's answer is incorrect. There is no specific solution to this problem that we are looking for. Y is not a defined value, x is not a defined value, and a is not a defined value. The -8 part is a constant and can easily be ignored. The only way you aren't getting an answer is by using an imaginary number unless I am reading the equation incorrectly.

Example:

y = Sqrt[-5]*8^2 - 8

Answer: -8 + 64 * i * Sqrt[5]

Simplified: -8 + 143.108 * i

i denotes the imaginary unit.

Thanks! That actually makes sense! I really appreciate you helping me and explaining the answer.

Thank you.

so you are saying it IS 4?

Sure. It can be any real number. 4, 24, or even 283,192.938 would all work. The only value of "a" that can **** this up is something that is imaginary since "a" is just a coefficient for "x".


You plug in 4:

y = 4x^2 - 8

Doesn't make a difference since "x" and "y" don't have a specific value they need to be.
If x is 0, y is -8.
If x is 1, y is -4.
If x is pi, y is 31.478
If x is infinity, y is infinity.

It really doesn't matter. "a" only ****s this up by being an imaginary number.

The question seems a little vague. "Has solutions" can mean a few things. For example let's say that you take it to mean "find values of x where y = 0", which is a common thing to find in algebra. Substitute 0 for y and solve for x and you'll get x = +/- sqrt(8/a). For a basic algebra class you could say that a <= 0 is bad.

If this is a homework question it's always a good idea to think of something similar that could be asked on a test. For example:

For what values of a does the following equation have no roots?
y = 8x^2 + a

i know all this, thus "a" cannot be 4 and the equation have no solutions.

therefore i can say about the value of "a", "it is not 4"

i agree that the problem is stupid.

Yeah, I thought it was really damn vague considering it's supposed to be a basic algebra problem. You would think it would be something as you described like finding a value that would cause the equation to not be equal.

I'm not following. Why would 4 cause it to have no solutions? As it stands, any real number for "a" will not cause any issues with unset values of "x" and "y".

it seems like he is missing a second equation and is supposed to be able to solve for x and y

This was my initial thought too. OP, you sure you posted all relevant material?

If so, your professor might be dumb.

Yeah, I don't think I learned about imaginary numbers until precalc/trig, but that was many years ago in HS.

No, I copied this EXACTLY. There isn't any other part to it.

a is infiniti. cuz if you subtract 8 from infiniti, you still have infiniti- and a ≠ y. do you see it now bro?