Thread: Doubt in non rational equation

Have a look at this irrational equation

Its domain is [1;+infty).
The main math packages (Derive, Maple and Mathematica) we have here, at school, give x = 0 and x = 2, both as solutions!
Actually if you square twice, at last you get:
x² - 2x = 0
that leads to x = 0 and x = 2, but x = 0 IMHO is not a solution...

What do you think about?

best regards

The "domain" mentioned assumes real numbers only. Apparently the packages mentioned assumes complex numbers.

When x=0:

sqrt(-1) - sqrt(-1) = sqrt(0)
i - i = 0
0 = 0 ...so x=0 satisfies the equation.

If you limit yourself to real numbers, then indeed x=0 is not a solution since sqrt(-1) is not a real number. If one squares both sides of the equation twice as mentioned, then solves the resulting quadratic, x=0 is considered an extraneous solution, unless complex solutions are assumed.

Darrell

Yes, x=0 is not a solution because the given domain of x is from 1 to infinity only.
Zero is not in the domain of x.

Mathcad (v.8 and 9) has the same funny behavior.

Try this: Ask Maple to solve

sqrt(5x-2) - sqrt(x-1) = sqrt(2x). (The only solution in the domain [1, infty) is x = 1/2 + sqrt(2)/2).

Mathcad returns this solution plus the solution x = 1/2 - sqrt(2)/2 and believes that each solution has multiplicity four!