When I solve the equation $(x-1) \cdot \sqrt{x^2 - 4}=0$ in the set of all real numbers (I have not known about complex numbers). I do following steps.

First step. I solve the inequality $x^2 - 4x \geqslant 0$, I have $x \leqslant -2 \lor x \leqslant$ 2.$

Second step. I solve the given equation by solving $x - 1 = 0$ and $x^2 - 4 = 0$. I get $x = -2 \lor x = 1 \lor x = 2.$

Third step. Check the above roots, we have the roots of the given equation are $x = -2$ or $x = 2.$

Thus, the given equation have two roots $x = -2$ or $x = 2.$

There are some one say that $x = 1$ is also a real root. Therefore the given equation have there roots $x = -2$, $x = 2$, and $x = 1$.

I think that, When $x=1$, the expression $\sqrt{x^2-4}= \sqrt{-3}$. If we don't know about complex numbers, how to understand $\sqrt{-3}$?

I think, solve the given equation in Reals domain is different from find all real numbers satify the given equation.

Is my solution correct when I solve in Reals domain?

I used Mathematica to check.

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And I asked my question at here https://www.mapleprimes.com/questions/227446-How-Many-Solutions-Are-There-In-The-Equationx1

Even, Maple has a bug!

  • $\begingroup$ I don't see the need to work with inequalities. This equation is equivalent to $(x-1)\sqrt{x-2}\sqrt{x+2}=0$, so the solutions are when $x-1=0$, $\sqrt{x-2}=0$, or $\sqrt{x+2}=0$. $\endgroup$
    – Dave
    Jul 10, 2019 at 2:01
  • $\begingroup$ Your answer is correct as you have found all real numbers satisfying the equation. $\endgroup$
    – Z Ahmed
    Jul 10, 2019 at 2:02
  • $\begingroup$ @Dave Are you sure your solution is correct? If we consider in Real number, then $\sqrt{x^2-4} \neq \sqrt{x-2} \sqrt{x+2}.$. $\endgroup$ Jul 10, 2019 at 4:25
  • $\begingroup$ Whether you know it or not, $ab = 0 \implies a = 0\ or\ b=0$ is valid in the complex numbers as well as the real numbers (that's because both the real numbers and the complex numbers satisfy the field axioms). So if the first factor is 0, then the product is 0, whether the second factor is real or complex. Moreover, the first factor is 0 when $x = 1$, so that is a perfectly good real solution to the equation. $\endgroup$
    – NickD
    Jul 10, 2019 at 5:46

1 Answer 1


This problem is solved by using the Zero Factor Property from algebra; i.e., $ab = 0$ iff $a = 0$ or $b=0$.

Hence, we have $x - 1 = 0$ or $x^{2} - 4 = (x+2)(x-2) = 0$; giving

$$ x = 1; x = \pm 2 $$

Remark1: $x=1$ causes the second term to be imaginary; yet, as a commentator points out, is a real solution.

Therefore, the solution set is $$ x = 1; \pm 2.$$

Remark2: $\sqrt{x^{2}-4} = 0$ iff $x^{2}-4 = 0.$

  • $\begingroup$ If I have not known about complex number, I think, the solution x = 1 is not a solution. $\endgroup$ Jul 10, 2019 at 3:51
  • $\begingroup$ When x=1, the expression \sqrt{x^2-4}= \sqrt{-3}. If you don't know about complex numbers, how to understand ` \sqrt{-3}`? $\endgroup$ Jul 10, 2019 at 4:04
  • $\begingroup$ @minhthien Thank you. $\endgroup$
    – I. Chekhov
    Jul 10, 2019 at 4:15
  • $\begingroup$ I don't vote for you. Because you did not read my solution carefully. $\endgroup$ Jul 10, 2019 at 4:16
  • $\begingroup$ Only Dr Zafar Ahmed DSc'comment is enough for me. $\endgroup$ Jul 10, 2019 at 4:19

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