Irrational Equation In Complex Mode

I've tried to solve this simple equation with many CAS:
sqrt(5*x)=sqrt(x-4)
Many software (see TI-Nspire and Maxima) in real mode give the wrong result (x=-1 that is not acceptable).
The Classpad give the right answer "No solution".

Here is my question.
For the same equation in complex mode am I allowed to accept the x=-1 solution? Classpad give the same "No solution" also in complex mode.
I've not understand the complex numbers or is a Classpad problem?

Thanks!

Edited by squonk, 13 November 2008 - 02:47 PM.

What do you mean x=-1 is unacceptable? You have to applaud the calculators that gave you that answer(they have superior CAS as far as dealing with such equation is concerned.)..By the way, what makes you think those calculators which gave you x=-1 are wrong?

And yes you're allowed to acept x=-1 and that classpad is wrong.Just think about it and try to solve it even without using a calculator(it's easy to see that the classpad which you used is damn wrong)
Goodluck

Edited by 2072, 24 November 2008 - 03:48 AM.

Thanks McCoy for your answer.
I've solved the equation FIRST with a pen and a paper sheet. The equation that I've posted has no solution in Real set because if you try to substitute the x with -1 value in the initial equation both square root assume the -5 value, and in Real sqrt(-5) don't exists (what's the number that got -5 as square?).
I don't think that I'm wrong here, don't you think?
But in complex I think that sqrt(-5) is defined, so I think that x=-1 is acceptable. If so the Classpad give the wrong answer...
Here I'm not so sure: this is why my question.

Edited by 2072, 24 November 2008 - 03:49 AM.
useles quote removed

Hi,
The ClassPad gives the right answer; x=1 in complex mode and “no solution” in real mode. If your OS is 3.03, otherwise it gives “no solution” in both cases if the OS is lower version.

Thank you for the answer.
I've not upgraded yet my Classpad. I make this soon.
Thanks again.

Thanks McCoy for your answer.
I've solved the equation FIRST with a pen and a paper sheet. The equation that I've posted has no solution in Real set because if you try to substitute the x with -1 value in the initial equation both square root assume the -5 value, and in Real sqrt(-5) don't exists (what's the number that got -5 as square?).
I don't think that I'm wrong here, don't you think?
But in complex I think that sqrt(-5) is defined, so I think that x=-1 is acceptable. If so the Classpad give the wrong answer...
Here I'm not so sure: this is why my question.

No problem at all Squonk.

yea if you use pen and paper for this type of question, substitution is the last thing to go for.First, square both sides .second subtract x from both sides.Third,divide both sides by 4....and you'll have your answer....no imaginary numbers involved.That's one of the many methods that those calculators used.Now the problem comes in if you want to verify by substituting x=-1 using a calculator.So for a calculator to 'see' that is true, it must use e^i(theta) =cos(theta) +i sin(theta) method, which is true for all real theta in radian.In this way, you can find the root of a negative number .Those calculators may have figured this out.

And yes you're not wrong,not even close to being wrong. x=-1 got -5 as a square and it's impossible to find roots of negative numbers directly, hence the complex mode.So if a calculator still gives you no solution even at this point, then honestly the calculator is desperately wrong.
have fun bro.

Edited by McCoy, 14 November 2008 - 06:59 AM.

No problem at all Squonk.

yea if you use pen and paper for this type of question, substitution is the last thing to go for.First, square both sides .second subtract x from both sides.Third,divide both sides by 4....and you'll have your answer....no imaginary numbers involved.That's one of the many methods that those calculators used.Now the problem comes in if you want to verify by substituting x=-1 using a calculator.So for a calculator to 'see' that is true, it must use e^i(theta) =cos(theta) +i sin(theta) method, which is true for all real theta in radian.In this way, you can find the root of a negative number .Those calculators may have figured this out.

And yes you're not wrong,not even close to being wrong. x=-1 got -5 as a square and it's impossible to find roots of negative numbers directly, hence the complex mode.So if a calculator still gives you no solution even at this point, then honestly the calculator is desperately wrong.
have fun bro.

LOL men i had to respond to your answer even though its a little old.... you try to sound like you know what youre saying but youre wrong!! How can you say substitution is the last thing to do?... in the minute you make square both sides youre asumming X as a real and in this of equation its NOT!!! Also its like saying the square of a negative number exists in real variable...

I'm sorry if it sounds a little harsh but youre wrong and the guy who asks its right... the Classpad is also right.. it gives No Solution in Real and X=-1 in Complex..