Algebra2
posted by Unknown .
Help?
1.2x^2/3+5^1/3=12
2. 6x^6/35x^1/36=0

#1
2x^(2/3) + 5x^(1/3) = 12 , you have a typo
let x^(1/3) = y
your equation becomes
2y^2 + 5y  12 = 0
(2y3)(y+4) = 0
y = 3/2 or y = 4
then x^(1/3) = 3/2  x = 27/8
if x^(1/3) = 4 > x = 64
check:
if x = 27/8
LS = 2(27/8)^(2/3) + 5(7/8)^(1/3)
= 2(9/4) + 5(3/2)
= 9/2 + 15/2 = 12 = RS
if x = 64
LS = 2(64)^(2/3) + 5(64)^(1/3)
= 2(16)  5(4)
≠ RS
x = 27/8
(in y = x^(1/3) , x would be defined as x>0
enter 2x^(2/3)+5x^(1/3)12 into "first graph" of
http://rechneronline.de/functiongraphs/
to see what I mean. There will be only one solution at x = 27/8 )

I think you have a typo in #2 if the question is supposed to be the same type
It should probably say
6x^(2/3)  5x^(1/3)  6 = 0
try #2 in the same way by letting y = x^(1/3)
it also factors, remember to reject a negative value of y