Please Explain How to Solve this Equation in Reals
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.
And I asked my question at here
Even, Maple has a bug!
$\endgroup$ 41 Answer
$\begingroup$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 $x^{2}-4 = 0.$
$\endgroup$ 10
