Question 637641: u^4/3- 17u^2/3 = -16
Answer by god2012(113)   (Show Source):
You can put this solution on YOUR website! Let U^1/3 = a.
therefore u = a^3.
Now , u^4/3- 17u^2/3 = u^1/3 * u - 17u^1/3 * u^1/3= -16
= a*a^3 - 17a*a = -16
= a^4 - 17a^2 = -16
= a^4- 17a^2 + 16 = 0
Let a^2 = x,
Now, x^2 - 17x +16 = 0 is a quadratic equation.
x^2 - 16x - x + 16 = 0
x (x -16) - 1(x - 16) = 0
(x-16)(x-1) = 0
x = 16 or x= 1
But x = a^2
so, a^2 = 16 or a^2 = 1
a = 4 or 1 = u^1/3
u= a^3 = 4^3 = 64 or 1
therefore u = 64 OR u = 1
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