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

Algebra2 
Reiny
#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
Respond to this Question
Similar Questions

Algebra2
Help ? 1. x3x^1/2+2=0 
Algebra2
If f = (1, 2), (2, 3), (3, 4), (4, 5), g = (1, 2), (3, 3), (5, 5), and h = (1, 0), (2, 1), (3, 2), Help? 
algebra2
5/21.(7/15) 
algebra2
5(6+2).2 
algebra2
2x+3=7 
algebra2
if a=5 and b=3 4a3b 
algebra2
s2/2s1/5=1/4 
Algebra2
Help? f(x)=5 ; x=3,2,1,0,1,2,3 
algebra2
(x5)^1/3 4=2 
Algebra2
23x58=8