Given the equation 9x4−1=0 Because equation degree is equal to = 4 - contains the even number 4 in the numerator, then the equation has two real roots. Get the root 4-th degree of the equation sides: We get: 494(1x+0)4=1 494(1x+0)4=−1 or 3x=1 3x=−1 Expand brackets in the left part
x*sqrt3 = 1
Divide both parts of the equation by sqrt(3)
x = 1 / (sqrt(3))
We get the answer: x = sqrt(3)/3 Expand brackets in the left part
x*sqrt3 = -1
Divide both parts of the equation by sqrt(3)
x = -1 / (sqrt(3))
We get the answer: x = -sqrt(3)/3 or x1=−33 x2=33
All other 2 root(s) is the complex numbers. do replacement: z=x then the equation will be the: z4=91 Any complex number can presented so: z=reip substitute to the equation r4e4ip=91 where r=33 - the magnitude of the complex number Substitute r: e4ip=1 Using Euler’s formula, we find roots for p isin(4p)+cos(4p)=1 so cos(4p)=1 and sin(4p)=0 then p=2πN where N=0,1,2,3,... Looping through the values of N and substituting p into the formula for z Consequently, the solution will be for z: z1=−33 z2=33 z3=−33i z4=33i do backward replacement z=x x=z
The final answer: x1=−33 x2=33 x3=−33i x4=33i