Posted by Unknown on Wednesday, August 29, 2012 at 5:03am.
#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
(2y-3)(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/function-graphs/
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
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