finding value of x

kenyu

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Aug 22, 2016
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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.
 
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.

No ...

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

x = ± 1 and/or ± i

You cannot avoid i in this case.
 
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.
It is worth noting that in the set of real numbers the 1\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 i\displaystyle \mathcal{i} to be a solution of x2+1=0\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 x41=0\displaystyle x^4-1=0 has four solutions: ±1, ±i\displaystyle \pm 1,~\pm\mathcal{i}.
 
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