Hi guys,

just a question please. This sum:

\(x^{2 \over3}-20=44\)

\(x^{2 \over3}=64\)

\(x^{({1 \over3})^2}=64\)

Let K = \(x^{1 \over3}\)

So \(K^2=8^2\)

Therefore K = 8

Sub back:

\(x^{1 \over3}=8\)

\(x^{1 \over3}=2^3\)

\(x=2^{3*3} \)

\(x=512\)

Tis is my reasoning. I am led to believe this is wrong and HAS to be done this way:

\(K^2-64=0\)

\((K-8)(k+8)=0\)

\(K=8\) or \(K=-8\)

\((x^{1 \over3})^3=8^3\) or \((x^{1 \over3})^3=-8^3\)

x=512 x= -512

Why must it be done this way?..why two answers?

juriemagic Apr 5, 2019

#1**0 **

x^(2/3) = 64 take sqrt of both sides

x^1/3 = + - 8 Cube both sides

x = 512 or -512

ElectricPavlov Apr 5, 2019

#2**+3 **

The only thing I can see wrong with your logic is that you that if

x^2 =64

x can be poitive or **negative** 8

other than that, your way is fine. :)

\(x^{2 \over3}-20=44\\ \left( x^{\frac{1}{3}}\right)^{2}=64\\ x^{\frac{1}{3}}=\pm8\\ x=8^3 \qquad or \qquad x=(-8)^3\\ x=512 \qquad or \qquad x=-512\\ \)

.Melody Apr 5, 2019

#3**+1 **

x^(2/3) - 20 = 44

x^(2/3) = 64

(x^2)^(1/3) = 64 cube both sides

x^2 = 64^3

x^2 = 262144 take both roots

x = ± √262144

x = ± 512

CPhill Apr 5, 2019

#4

#6**+3 **

x^(2/3) ⇒ (-512)^(2/3) = 64 (-1)^(2/3) ≈ -32. + 55.4256 i ?????

that is not correct

x^(2/3)

when x=-512

⇒ (-512)^(2/3)

= (64) *(-1)^(2/3) true

(-1)^(2/3) = [(-1)^2]^(1/3) = [1]^(1/3) = 1

or

(-1)^(2/3) = [ (-1)^(1/3) ]^(2) = [-1]^(2) = 1

either way

= (64) *(-1)^(2/3)

= +64

I am glas you showed us your confusion. People do not learn if they do not ask questions :)

Melody
Apr 5, 2019

#7**+3 **

yep yep....I have it...a square root ALWAYS has a positive and negative answer...my my...how could I forget that????.....Thank you to all who replied..I do appreciate..

juriemagic
Apr 5, 2019