Pool

If water flows into the pool by two inlets, fill the whole for 19 hours. The first inlet filled pool 5 hour longer than the second. How long pool take to fill with two inlets separately?

Correct result:

t₁ =  40.66 h
t₂ =  35.66 h

Solution:

$$\begin{gathered} {\displaystyle \frac{1}{t_1}}+{\displaystyle \frac{1}{t_2}}={\displaystyle \frac{1}{19}} \\ {t}_2=t_1-5 \\ 1\mathrm{/}{t}_1+1\mathrm{/}(t_1-5)=1\mathrm{/}19 \\ 19\ast (x-5)+19\ast x=(x-5)\ast x \\ 19\cdot (x-5)+19\cdot x=(x-5)\cdot x \\ -x^2+43x-95=0 \\ {x}^2-43x+95=0 \\ a=1;b=-43;c=95 \\ D=b^2-4ac=43^2-4\cdot 1\cdot 95=1469 \\ D>0 \\ {x}_{1,2}={\displaystyle \frac{-b\pm \sqrt{D}}{2a}}={\displaystyle \frac{43\pm \sqrt{1469}}{2}} \\ {x}_{1,2}=21.5\pm 19.163767896737 \\ {x}_1=40.663767896737 \\ {x}_2=2.3362321032632 \\ \text{ Factored form of the equation: } \\ (x-40.663767896737)(x-2.3362321032632)=0 \\ {t}_1>0 \\ {t}_1=x_1=40.6638=40.66\text{ h} \end{gathered}$$

Our quadratic equation calculator calculates it.

$$\begin{gathered} t_2>0 \\ {t}_2=t_1-5=40.6638-5=35.66\text{ h} \end{gathered}$$

Try another example

Showing 3 comments:

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
ubless i made a mistake, your calculations need reexamination!!!        Correct me, please.

2 years ago  2 Likes Like

Dr Math

right side of equation is wrong - should be t1*(t1-10) = t1² - 10*t1 now t1³-10t1

2 years ago  Like

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!!!!

2 years ago  Like

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