Ten Points, Two easy math questions?

Question

1)A speedboat takes 1 hour longer to go 24 miles up a river than to return. If the boat cruises at 18 miles per hour in still water, what is the rate of the current?

2)Solve the equation by introducing a substitution that transforms the equation to quadratic form.

t^(-2/3) - t^(-1/3) - 30=0

Give exact answers in form of fraction. Write the answers in ascending order.

Answer 1

1)

the time speedboat uses to go down river (return):

time = distance / speed

time = 24 / (18 + x) , where x is the waterspeed (rate) in river downstream

going up river it takes 1 hour more:

time + 1 = 24 / (18 - x)

24 / (18 + x)  + 1 = 24 / (18 - x)

(24 + 18 + x) / (18 + x) = 24 / (18 - x)

(42 + x) (18 - x) = 24 (18 + x)

756 + 18x - 42x - x^2 = 432 + 24x

324 - 48x - x^2 = 0

x1 = -54 mph (not valid since water speed is positive)

x2 = 6 mph

water current is 6 mph

2)

t^(-2/3) - t^(-1/3) - 30 = 0 , t^(-1/3) = u

u^2 - u - 30 = 0

u1 = -5

u2 = 6

t^(-1/3) = -5

(t^(-1/3))^-3 = -5^-3

t = -0,008

t^(-1/3) = 6

(t^(-1/3))^-3 = 6^-3

t = 0,0046