I know this question has been solved in here:

Factor $x^4 + 64$

But I solved in another way and I don't seem to be getting the same solution and I don't know where is the error:

Here is my attempt:

$x^4 + 64 = 0$
$x^4 = -64$
Let $m = x^2$
$m^2 = -64,$ take the square root on both sides
$m_1= +i8 , m_2 = -i8$, substitute x back:
$x^2 = i8$, take the square root on both sides
$x^2 = -i8$, take the square root on both sides
We get:
$x_1 = + \sqrt{i8}= 2 \sqrt{2}\sqrt{i}$,
$x_2 = + \sqrt{i8}= 2 \sqrt{2}\sqrt{i}$,
$x_3 = - \sqrt{-i8}= 2i \sqrt{2}\sqrt{i}$,
$x_4 = - \sqrt{-i8}= 2i \sqrt{2}\sqrt{i}$,

When I plug the equation in Wolf alpha it gave this result: $x = \pm (2 - 2 i)$, $x = \pm (2 - 2 i)$

Where is the mistake in my solution?

  • 6
    $\begingroup$ How do you write $\sqrt{i}$ in the form $a+ib$? $\endgroup$ – mickep Mar 2 '17 at 6:25
  • $\begingroup$ @mickep I am not sure how to do that $\endgroup$ – CS_XYZ Mar 2 '17 at 6:30
  • 1
    $\begingroup$ Then I must ask you what you mean by what you have written. After all, you were the one writing $\sqrt{i}$. $\endgroup$ – mickep Mar 2 '17 at 6:31
  • $\begingroup$ I followed a simple method of solving the polynomial but I got there .. did I make a mistake by writing it this way? $\endgroup$ – CS_XYZ Mar 2 '17 at 6:33
  • $\begingroup$ Try to solve the equation $(a + b i)^2 = i$. $\endgroup$ – Qudit Mar 2 '17 at 6:33

$i=\cos(\pi/2)+i\sin(\pi/2)$, so $\sqrt{i}=\cos(\pi/4)+i\sin(\pi/4)=\sqrt{2}(i+1)/2$ or $\cos(5\pi/4)+i\sin(5\pi/4)=-\sqrt{2}(i+1)/2$ depends on which branch you choose.

Do similar thing on $-i$, plug into your $\sqrt{i}$ and $\sqrt{-i}$, your answer should be the same as the one you get from Wolf alpha.

  • $\begingroup$ Thanks, I'm sorry if this would sound as dump question but if I plug $\sqrt(-i)$ then what exactly would change? + what is the name of the method you used? $\endgroup$ – CS_XYZ Mar 2 '17 at 6:53
  • $\begingroup$ It's Euler's Formula. $\endgroup$ – John Smith Mar 2 '17 at 6:55

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