SOLUTION: u^4/3- 17u^2/3 = -16

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

RELATED QUESTIONS
(17u^5y^3+10u^3y^2) /... (answered by Fombitz)

3x2+7x+2=0 3x2-16x+5=0 7x2-9x+2=0... (answered by checkley75)

6e+u=3... (answered by josgarithmetic)

solve for u:... (answered by Fombitz)

Factor completely U 4 W 3 - 16 w... (answered by rchill)

{{{((u^(3/4))(u^(1/3)))/((u^(2/3))(u^(5/6)))}}}=u^(a/b) where a=? and... (answered by venugopalramana)

u-3/4=6 2/3 (answered by stanbon)

Show steps rhank u... (answered by lynnlo)

Simplify the expression.... (answered by MathLover1,MathTherapy)