finding value of x

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kenyu

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I got this answer x^2 = -1 so x = sqrt of -1 which is imaginary number. The question is can I do this if x^2 = -1 I will squared both sides so x^4 = 1 then fourth root both sides so I will get x = positive or negative 1. No need for imaginaries.
 
kenyu said:
I got this answer x^2 = -1 so x = sqrt of -1 which is imaginary number. The question is can I do this if x^2 = -1 I will squared both sides so x^4 = 1 then fourth root both sides so I will get x = positive or negative 1. No need for imaginaries.
Click to expand...

No ...

x4 = -1 → Now x has 4 solutions instead of 2!!

x = ± 1 and/or ± i

You cannot avoid i in this case.
 
kenyu said:
I got this answer x^2 = -1 so x = sqrt of -1 which is imaginary number. The question is can I do this if x^2 = -1 I will squared both sides so x^4 = 1 then fourth root both sides so I will get x = positive or negative 1. No need for imaginaries.
Click to expand...
It is worth noting that in the set of real numbers the \(\displaystyle \sqrt{-1}\) does not exist. As a result the equation x^2=-1 has no solution. But the real number system has a model enlargement that includes a number that solves that difficulty.

Define \(\displaystyle \mathcal{i}\) to be a solution of \(\displaystyle x^2+1=0\). Now we have a new system called complex numbers that conforms to most of the usual properties of real numbers. (there are a few exceptions: we don't use the radical signs; exponents and logarithms require further restrictions, etc)

In your case as Mr. Khan has shown you the equation \(\displaystyle x^4-1=0\) has four solutions: \(\displaystyle \pm 1,~\pm\mathcal{i}\).
 
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