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 06-25-2012, 11:22 AM #1 enthusiast     Join Date: Apr 2012 Posts: 64 Need help with a basic algebra problem??? What can you say about the value of "a" if the equation y=ax^2-8 has no solutions?
 06-25-2012, 12:05 PM #2 DKGOAT     Join Date: May 2006 Location: post hard in da mother****ing paint Posts: 58,115 Re: Need help with a basic algebra problem??? a is not 4
 06-25-2012, 12:31 PM #3 enthusiast     Join Date: Apr 2012 Posts: 64 Re: Need help with a basic algebra problem??? Ok, thank you very much. I really appreciate it. Would you mind explaining how you came to this conclusion please?
 06-25-2012, 02:33 PM #4 Pooh-Bah     Join Date: Mar 2011 Posts: 4,488 Re: Need help with a basic algebra problem??? 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. Last edited by GusJohnsonGOAT; 06-25-2012 at 02:43 PM.
06-25-2012, 02:48 PM   #5
enthusiast

Join Date: Apr 2012
Posts: 64
Re: Need help with a basic algebra problem???

Quote:
 Originally Posted by GusJohnsonGOAT 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.

06-25-2012, 03:06 PM   #6
DKGOAT

Join Date: May 2006
Location: post hard in da mother****ing paint
Posts: 58,115
Re: Need help with a basic algebra problem???

Quote:
 Originally Posted by GusJohnsonGOAT dkgo's answer is incorrect.
so you are saying it IS 4?

06-25-2012, 03:27 PM   #7
Pooh-Bah

Join Date: Mar 2011
Posts: 4,488
Re: Need help with a basic algebra problem???

Quote:
 Originally Posted by dkgojackets 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.

Last edited by GusJohnsonGOAT; 06-25-2012 at 03:33 PM.

 06-25-2012, 03:46 PM #8 Pooh-Bah     Join Date: Dec 2003 Location: US Posts: 3,618 Re: Need help with a basic algebra problem??? 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
06-25-2012, 03:53 PM   #9
DKGOAT

Join Date: May 2006
Location: post hard in da mother****ing paint
Posts: 58,115
Re: Need help with a basic algebra problem???

Quote:
 Originally Posted by GusJohnsonGOAT 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.
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.

06-25-2012, 04:12 PM   #10
Pooh-Bah

Join Date: Mar 2011
Posts: 4,488
Re: Need help with a basic algebra problem???

Quote:
 Originally Posted by jmark 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

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.

Quote:
 Originally Posted by dkgojackets 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.
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".

 06-25-2012, 04:15 PM #11 DKGOAT     Join Date: May 2006 Location: post hard in da mother****ing paint Posts: 58,115 Re: Need help with a basic algebra problem??? it seems like he is missing a second equation and is supposed to be able to solve for x and y
06-25-2012, 04:16 PM   #12
Carpal \'Tunnel

Join Date: Aug 2008
Location: learnin how to dougie
Posts: 7,516
Re: Need help with a basic algebra problem???

Quote:
 Originally Posted by dkgojackets 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.

06-25-2012, 04:18 PM   #13
Pooh-Bah

Join Date: Mar 2011
Posts: 4,488
Re: Need help with a basic algebra problem???

Quote:
 Originally Posted by Vintage00 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.

06-25-2012, 05:44 PM   #14
enthusiast

Join Date: Apr 2012
Posts: 64
Re: Need help with a basic algebra problem???

Quote:
 Originally Posted by Vintage00 This was my initial thought too. OP, you sure you posted all relevant material? If so, your professor might be dumb.
No, I copied this EXACTLY. There isn't any other part to it.

 06-25-2012, 09:24 PM #15 banned   Join Date: Jan 2010 Location: what would trevor bauer do Posts: 2,966 Re: Need help with a basic algebra problem??? a is infiniti. cuz if you subtract 8 from infiniti, you still have infiniti- and a ≠ y. do you see it now bro?

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