$-\sqrt{-3} * \sqrt{-3} = x$

$-(-3^{\frac{1}{2}}) * (-3^{\frac{1}{2}}) = x$

$ +3^{\frac{1}{2}} * -3^{\frac{1}{2}} = x$

$-3^-1 = x$, then $x = -3$

The correct result should be $3$ positive

And i can get that result with second development:

$-(\sqrt{3}i) * (\sqrt{3}i) = x$

$-\sqrt{3}i * \sqrt{3}i = x$

$-\sqrt{9}i^2 = x$

$-3 i^2 = x$, then $x = 3$

I want to know, what I am wrong with in the first way to solve it (using powers), I still do not have major knowledge of the school, therefore I need an elementary explanation, I have read some similar questions, but I have not understood the answers, in advance thanks .

  • 3
    $\begingroup$ There may be a lot of reasons (such as roots are multivalued), but the most obvious reason is that $\sqrt{-1}=i$, whereas in your calculations, you take $\sqrt{-1}=-1$. $\endgroup$ – Clayton May 1 '18 at 22:04
  • $\begingroup$ Arithmetic can be misleading on $\Bbb C$. Note that $$-(-3)^{1/2}=(-1)^{3/2}3^{1/2}\neq +3^{1/2}$$ There's a factor of $(-1)^{1/2}$ involved which is defined as the imaginary unit $i$. $\endgroup$ – Prasun Biswas May 1 '18 at 22:05
  • $\begingroup$ Why $(-1)^{\frac{3}{2}} * 3^{\frac{1}{2}}$ instead of $-[(-1)^{\frac{3}{2}} * 3^{\frac{1}{2}}]$, where is the $-$ of the $(-3)^{\frac{1}{2}}$ $\endgroup$ – Mattiu May 1 '18 at 22:09
  • $\begingroup$ @Clayton how?, can you explain me, please $\endgroup$ – Mattiu May 1 '18 at 22:12
  • 1
    $\begingroup$ $\sqrt{-3} = (-3)^{\frac 12} \ne (-3^{\frac 12}) = -(3^{\frac 12})$. $\endgroup$ – fleablood May 1 '18 at 22:18

$$-\sqrt{-3}=-(-3)^\frac{1}{2}\ne3^{\frac{1}{2}}$$ That would be the same as saying $$-i\sqrt{3}=\sqrt{3} \rightarrow i=-1$$ Which is obviously false.

It's not possible to solve your equation in real terms because: $$\sqrt{a}\sqrt{b}=\sqrt{ab}\iff a,b\ge0$$ And $a$ and $b$ are both $-3$ here, which clearly isn't $>0$.

Instead, like you did, use complex numbers to solve this


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