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Question: How do you find all the real and complex roots of \({{x}^{4}}=1\) ? (a) Factorization (b) Guess...

How do you find all the real and complex roots of x4=1{{x}^{4}}=1 ?
(a) Factorization
(b) Guessing the roots
(c) Changing the variable
(d) None of the above

Explanation

Solution

To find all the real and complex roots, we are to try to factorize the equation given as, x4=1{{x}^{4}}=1. We are going to analyze the factors of the equation and then try to configure each of them equaling to zero. As it is a 4th degree equation, we are going to find the 4 roots as ±1,±i\pm 1,\pm i, from this given equation, may it be real and complex.

Complete step-by-step answer:
According to the question, we have been asked to find the methods of finding roots of a cubic equation.
We are trying to factorize the equation, f(x)=x41f(x)={{x}^{4}}-1,
Now, to start with,
x4=1{{x}^{4}}=1
x41=0\Rightarrow {{x}^{4}}-1=0
as a2b2=(a+b)(ab){{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right) , we get
(x2+1)(x21)=0\Rightarrow ({{x}^{2}}+1)({{x}^{2}}-1)=0
By factorizing, we are getting, for i2=1{{i}^{2}}=-1 ,
(x2i2)(x21)=0\Rightarrow ({{x}^{2}}-{{i}^{2}})({{x}^{2}}-1)=0
Thus, with more simplification we get,
x21=0{{x}^{2}}-1=0
(x)2(1)2=0\Rightarrow {{(x)}^{2}}-{{(1)}^{2}}=0
as a2b2=(a+b)(ab){{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right) , we get
(x+1)(x1)=0\Rightarrow (x+1)(x-1)=0
So, we get, x=1,1x=1,-1 .
These are the real roots of the given equation x4=1{{x}^{4}}=1.
And again,
x2i2=0{{x}^{2}}-{{i}^{2}}=0
as a2b2=(a+b)(ab){{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right) , we get
(x+i)(xi)=0\Rightarrow (x+i)(x-i)=0
x=i,i\Rightarrow x=i,-i
So, we are getting, x=i,ix=i,-i
These are the imaginary roots of the equation x4=1{{x}^{4}}=1.
So, all the roots of this equation is said to be, ±1,±i\pm 1,\pm i .
And we can also check that, the hit and trial method can be used to find the solutions of our given problem.
To describe the hit and trial method,
Let us try with the value -1 to start with,
So, we have our equation as, x41{{x}^{4}}-1,
putting -1 in the value of x, we get,
(1)41\Rightarrow {{\left( -1 \right)}^{4}}-1
Now, simplifying the value,
(1)21\Rightarrow {{\left( 1 \right)}^{2}}-1
After more simplification, we will reach the value we want to evaluate.
11=0\Rightarrow 1-1=0
So, we are getting the value of the function as 0 by putting -1. Then we can conclude that -1 is the root of the equation.
Similarly if we put the values of x as 1,±i1,\pm i we will also get the values of the function as zero. Then we can also conclude that the 1,±i1,\pm iare the roots of the equation.
Hence, the correct option is, (a) Factorization and (b) Guessing the roots.

Note: In relation to quadratic equations, imaginary numbers (and complex numbers) occur when the value under the radical portion of the quadratic formula is negative. When this occurs, the equation has no roots (zeros) in the set of real numbers. The roots belong to the set of complex numbers, and will be called "complex roots" (or "imaginary roots"). These complex roots of the equation x4=1{{x}^{4}}=1 can be expressed in the form a + bi, with the known fact that, i2=1{{i}^{2}}=-1 . On the other hand, putting the value of x2{{x}^{2}} as t would give us a quadratic equation and the problem can be solved easily by that.