# Pool

If water flows into the pool by two inlets, fill the whole for 8 hours. The first inlet filled pool 6 hour longer than second. How long pool take to fill with two inlets separately?

Result

t1 =  19.54 h
t2 =  13.54 h

#### Solution:

$\dfrac{ 1 }{ t_{1} } +\dfrac{ 1 }{ t_{2} }=\dfrac{ 1 }{ 8 } \ \\ t_{2}=t_{1} - 6 \ \\ 1/t_{1} + 1/(t_{1}-6)=1/8 \ \\ \ \\ 8*(x-6) + 8*x=(x-6) * x \ \\ \ \\ 8 \cdot \ (x-6) + 8 \cdot \ x=(x-6) \cdot \ x \ \\ -x^2 +22x -48=0 \ \\ x^2 -22x +48=0 \ \\ \ \\ a=1; b=-22; c=48 \ \\ D=b^2 - 4ac=22^2 - 4\cdot 1 \cdot 48=292 \ \\ D>0 \ \\ \ \\ x_{1,2}=\dfrac{ -b \pm \sqrt{ D } }{ 2a }=\dfrac{ 22 \pm \sqrt{ 292 } }{ 2 }=\dfrac{ 22 \pm 2 \sqrt{ 73 } }{ 2 } \ \\ x_{1,2}=11 \pm 8.5440037453175 \ \\ x_{1}=19.544003745318 \ \\ x_{2}=2.4559962546825 \ \\ \ \\ \text{ Factored form of the equation: } \ \\ (x -19.544003745318) (x -2.4559962546825)=0 \ \\ \ \\ t_{1}>0 \ \\ t_{1}=x_{1}=19.544 \doteq 19.544 \doteq 19.54 \ \text{h}$

Checkout calculation with our calculator of quadratic equations.

$t_{2}>0 \ \\ t_{2}=t_{1} - 6=19.544 - 6=\dfrac{ 677 }{ 50 }=13.54 \ \text{h}$

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Leave us a comment of this math problem and its solution (i.e. if it is still somewhat unclear...):

Math student
1/t1+1/(t1-10)=1/18
multiply each term by18(t1)(t1-10)
that results in
18(t1-10)+18t1=t1(t1)(t1)-10t1
using the quadratic formula results in t1=-49.6 and 3.63

1 year ago  2 Likes
Dr Math
right side of equation is wrong - should be t1*(t1-10) = t12 - 10*t1 now t13-10t1

Math student
the problems seems to have changed - - - t2 is now equal t1-6

therefore 1/t1+1/(t1-6)=1/18
multiplying each term by18(t1)(t1-6) ==== 18(t1-6)+18t1=t1(t1-6), simplifying further 18t1-108+18t1=t12-6t1
or 0=t12-6t1-18t1+108
graphing y=18(t1-6)+18t1-t1(t1-6) results in t1=39.25 hours and t2=39.25-6=33.25 hours (same as your NEW answer!!!!

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