Algebra II question 100m8 + -9 = 0 solve for m when m is to the power of 8

Question

100m8 + -9 = 0

Solve for m using factorization method

Answer 1

100m⁸-9 = (10m⁴+3)(10m⁴-3)=0

m = ±⁴√(3/10) , ±i ⁴√(3/10)

Answer 2

100m^8 - 9 = 0

(10m^4)^2 - (3)^2 = 0

factor as the difference of two squares: a^2-b^2 = (a-b)(a+b)

(10m^4 -3)(10m^4 +3) = 0

set each factor = 0, solve for m

10m^4 = 3

m^4 = 3/10 = 0.3

m^2 = + or - sqr(.3) = about + .55

m = sqr.55 = about 0.74 or -0.74

plus two other imaginary solutions, + .74i. but an 8th degree equation has 8 solutions. so there're 6 imaginary solutions altogether

100m^8 = 9

10m^4 = + or - 3

m^4 = + or - .3

m^2 = + or - sqr.3 and + or - isqr(-3)

m = + or - the 4th root of 0.3 which are 2 real numbers and 2 imaginary numbers

or m = + or - the 4th roots of -0.3 which are all imaginary

if you graphed the equation 100m^8 -9 = 0

it would have 2 x intercepts, 7 possible turning points, all above the x axis

Using Descartes Rule of Signs there is only one change in sign from the 100m^8 term to the constant term, so there is a maximum one positive real root. Replacing m by -m also has only one change in sign, so there is a maximum one negative real root.