I have an Advanced Algebra problem: write all real roots of x^3=125 as a decimal and x^4=10000 as a decimal.!

Question

Also evaluate 125 to the 1/3 power and 144 to the 1/2 power!

Also evaluate 125 to the 1/3 power and 144 to the 1/2 power and -(49) to the 1/2 power!

Answer 1

Seeing as this is ADVANCED, I'd think our answers are lacking some substance.

1. x^3 = 125
x = 5

x^3 - 125 = (x-5)(x^2+5x+25) = 0

x^2 + 5x + 25 = 0
[ -5 +/- sqrt (25 - 100) ] / 2 = x_1,2
[ -5 +/- 5i sqrt 3 ] / 2 = x_1,2

Perhaps you want it in this form.
-5 [ 1/2 +/- (i sqrt 3) / 2 ] = x_1,2
-5 [ cos 60 +/- i sin 60 ] = x_1,2

x^4 = 10000
x = 10 or -10

x^4 - 10000 = (x^2-100)(x^2+100)
x^2 + 100 = 0
x^2 = -100
x = 10i or -10i

Answer 2

x^3=125

x = 5.0000000000000000000000000000000000000000000...
(round off to the necessary place value_


x^4=10000

x = -10.0000 or x = 10.0000


125 to the 1/3 power is 5 since 5^3 = 125

144 to the 1/2 power is 12 since 12^2 = 144


[another answer gives 144^(1/2) as 12 or -12. But -12 is incorrect. Even though (-12)^2 = 144, by definition 144^(1/2) cannot be negative.]

Answer 3

x^3-125 is the difference of two cubes x^3 and 5^3 and so the equation x^3-125 = 0 has only one real root x = 5

Note x^3-125 = (x - 5)*(x^2 + 5x + 25)

x^4 - 10000 = (x^2 - 10^2)*(x^2 + 10^2)
etc

Answer 4

x^3 = 125
x = 5.0

x^4 = 10000
x = 10.0, -10.0

125^1/3 = 5.0

144^0.5 = 12.0, -12.0