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.**