Question 918875
Two campers left the camp at 8:00 AM to paddle their canoe 10 Km upstream to a fishing cove.  They finished for 3.5 hours then paddled back to the campsite and arrived 6:00 PM.  If the rate of the current was 3 km/hr. How fast could the camper paddle in still water?


let the rate of boat in still water be x

upstream time = 10/(x-3)

Time spent there = 3.5 hours

down stream time = 10/(x+3)


They left at 8.00 and reached at 6.00 pm

=10 hours


10/(x-3)+3.5+10/(x+3) = 10

10/(x-3)+10/(x+3) = 10-3.5

10/(x-3)+10/(x+3) = 6.5

take the LCM 

10(x+3)+10(x-3)= 6.5(x+3)(x-3)

10x+30+10x-30=6.5(x^2-9)

20x=6.5x^2-58.5


6.5x^2-20x-58.5=0

Find the roots of the equation by quadratic formula							
							
a=	6.5	,	b=	-20	,	c=	-58.5
							
b^2-4ac=	400	+	1521				
b^2-4ac=	1921						
{{{	sqrt(	1921	)=	43.83	}}}		
{{{x=(-b+-sqrt(b^2-4ac))/(2a)}}}							
{{{x1=(-b+sqrt(b^2-4ac))/(2a)}}}							
x1=(	20	+	43.83	)/	13		
x1=	4.91						
{{{x2=(-b-sqrt(b^2-4ac))/(2a)}}}							
x2=(	20	-43.83	) /	13			
x2=	-1.83						
Ignore negative value							
boat   	speed	4.910	km/h