Hello, would someone kindly answer this question?

Question

 In the event of a power failure, a computer model estimates the temperature T (in degrees 

    Celsius) of a food processing plant’s freezer to be Tt= ((2t^2)/(t+2) ) -4, where t is the time in hours 

    after the power failure. If the power failure happened at 4pm, find the time when the ice cubes 

    in the freezer started to melt. 

 

Answer 1

Ice cubes melt at temperature above 0 degree celsius.

Or  ((2t^2)/(t+2) ) -4 =0

Or 2t^2 = 4t + 8 =0

Or 2t^2 - 4t -8 = 0

or t^2 -2t - 4 =0

On solving the following quadratic equation using Quadratic formula, t = -b +- sqrt(b^2 -4ac)/2a, we get:

t =  1 - sqrt(5)  or   t = 1 + sqrt(5)

Neglecting t = 1- sqrt(5) as time doesnot flow backward.

Therefore, the ice cubes will start melting at  4 + 1 + sqrt(5) = 5 + sqrt(5)

or it will start melting at 08: 14 : 10 pm

 

Answer 2

Ice cubes melt at 0°C so:

0=2t²/(t+2)-4,

t²/(t+2)=2,

t²=2t+4,

t²-2t=4,

t²-2t+1=5,

(t-1)²=5,

t-1=±√5,

t=1+√5 (reject the negative result),

t=3.236hr approx after 4pm.

The ice cubes will melt around 7:14pm.