+1124 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?
+36916 x^(2/3) = 64 take sqrt of both sides
x^1/3 = + - 8 Cube both sides
x = 512 or -512
+118608 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. :)
\(x^{2 \over3}-20=44\\ \left( x^{\frac{1}{3}}\right)^{2}=64\\ x^{\frac{1}{3}}=\pm8\\ x=8^3 \qquad or \qquad x=(-8)^3\\ x=512 \qquad or \qquad x=-512\\ \)
+128408 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
+118608 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 :)
+1124 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..
