anonymous
  • anonymous
HOW TO find the imaginary solution for x^4+1=0?? pLEASE HELP....
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
To get it I would factor it into (x^2+1)(x^2-1) [sum of perfect squares] and then factor these to (x+1)(x-1)(x+i)(x-i) so your imaginary solutions are i and -i
anonymous
  • anonymous
it is not factorable , I thought
anonymous
  • anonymous
oops I just realized my first factors are in fact incorrect. I misread the original as x^4-1

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More answers

anonymous
  • anonymous
x^4=-1 x=fourth root of -1
anonymous
  • anonymous
the answer is 1+i/square root of 2,, but I dont know why
sgadi
  • sgadi
Since equation is of order 4, we get four different solutions in complex domain. 1+0i -1+0i 0+1i 0-1i or simply \[\pm i\mbox{ , }\pm 1\]
anonymous
  • anonymous
square root of -1 is i square root of i is -i and i
anonymous
  • anonymous
Instead factor it to (x^2 + i)(x^2 - i)
anonymous
  • anonymous
yes.. I did, but ten what's square root of I ??
anonymous
  • anonymous
So from the quadratic formula we get (1+√i)/2, (1-√i)/2, (-1+√i)/2, and (-1-√i)/2 but as for √i..... I suppose you could rewrite it as (-1)^{1/4}
sgadi
  • sgadi
please see my answer above ... those are roots of 1.
anonymous
  • anonymous
I agree that \[\pm{i}\] are solutions but 1 and -1 are not. If you go back to the original problem and substitute them in you get an incorrect statement. 1^4 + 1≠0
sgadi
  • sgadi
I am sorry ... I have done a terrible mistake in the previous calculation. given equation \[x^4=-1\] since \[e^{i\theta}=\cos{\theta}+i\sin{\theta}\] So, we can write \[-1=e^{i\pi}\] So, we can say \[x=(-1)^{1/4}=e^{\frac{i\pi}{4}}=\cos{\frac{\pi}{4}}+i\sin{\frac{\pi}{4}}\] \[\frac{1}{\sqrt{2}}+i\frac{1}{\sqrt{2}}\] Similarly taking diffrent values of \[\theta\] we can get \[\frac{1}{\sqrt{2}}-i\frac{1}{\sqrt{2}}\] \[-\frac{1}{\sqrt{2}}+i\frac{1}{\sqrt{2}}\] \[-\frac{1}{\sqrt{2}}-i\frac{1}{\sqrt{2}}\] are also roots for the given equation.

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