Lily swims for 5 hours in stream that has a current of 1 mph. She leaves her dock, swims upstream for 2 miles and then swim back to her dock. What is her swimming speed in still water?

Answer:Swimming speed in still water is approximately 1.5 mphStep-by-step explanation:Given the speed of the current = 1 mph distance she swims upstream =2 miles, the total time = 5 hours. She swam 2 miles upstream against the  current and 2 miles back to the dock with  the current. The formula that relates  distance, time, and rate is [tex]d = r t\ \ \ \ or  \ \ \  t= \frac{d}{r}[/tex]Let x be the speed in still water.Then  her speed with the current is x + 1, and her speed against the current is x – 1.Total time is equal to 2 miles with upstream and 2 miles downstream.[tex]\frac{2}{x+1}+\frac{2}{x-1}=5\\\\\frac{2(x-1)}{(x+1)(x-1)}+\frac{2(x+1)}{(x-1)(x+1)}=5\\\\\frac{2x-2}{(x^2-1)}+\frac{2x+2)}{(x^2-1)}=5\\\\\frac{2x-2+2x+2)}{(x^2-1)}=5\\\\4x=5(x^2-1)\\4x= 5x^2-5\\0= 5x^2-4x-5[/tex]Now Using quadratic formula to solve above equation we get;[tex]x= \frac{-b \±\sqrt{b^2-4ac}} {2a}\\\\here \ \ a= 5,b=-4,c=-5\\\\x= \frac{-(-4) \±\sqrt{(-4)^2-4\times 5\times-5}} {2\times5}\\\\x=\frac{4 \±\sqrt{16+100}} {10}\\\\x=\frac{4 \±\sqrt{116}} {10}=x=\frac{4 \±2\sqrt{29}} {10}= x=\frac{2 \±\sqrt{29}} {5}\\ x \approx 1.5 \ or \ x = -0.7 [/tex]Since speed must be positive, Hence speed of still water is about 1.5 miles per hour.